Chapter 7.7- Velocity of the Center of mass

In the previous section, we saw the method for finding the location of 'C' of any system of particles. In this section we will see the applications of 'C'

1. Let a system consist of two particles P and Q
• Let both P and Q be in motion
2. Consider any instant during that motion
• In earlier sections of this chapter, we have seen that, all particles in the system need not be having the same velocity at an instant (Details here)
• So let us assume that, at that instant, P and Q are having different velocities
3. Consider the instant at which the reading in the stop-watch is 't₁'
    ♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't₂'
    ♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t2)}}$
4. If we subtract the 'initial position vector' from the 'final position vector', we will get the displacement (Details here)
• So we can write:
The displacement suffered by 'P' during the time interval of (t₂-t₁) 
= $\mathbf\small{\vec{\Delta r}_P=\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}$
5. If we divide the displacement by the 'time interval during which the displacement took place', we will get the average velocity (Details here)
• Time interval during which the displacement took place = (t₂-t₁) = Δt
• So we can write:
Average velocity with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{\Delta r}_P}{\Delta t}=\frac{\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{r}_{P(t2)}}{\Delta t}-\frac{\vec{r}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average velocity with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{Q(\Delta t)}=\frac{\vec{r}_{Q(t2)}}{\Delta t}-\frac{\vec{r}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above velocities with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}=\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}$
• This is $\mathbf\small{\sum{m_i \vec{r}_{i(t2)}} }$
• That means: 
(i) The position vectors of all the particles at the 'instant when reading of the stop-watch is t₂' is taken
(ii) The summation is done using those position vectors
10. But we have Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The summation in the numerator of Eq.7.4 is the same summation that we wrote in (9)
11. Now, Eq,7.4 can be written as: $\mathbf\small{M\,\vec{r}_C=\sum{} \,m_i\,\vec{r}_i}$
• So the summation in (9) can be replaced by $\mathbf\small{M\,\vec{r}_{C(t2)}}$
Where $\mathbf\small{\vec{r}_{C(t2)}}$ is the position vector of 'C' at the 'instant when reading of the stop-watch is t₂'
12. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$ 
13. The 'numerator of the second term' on the right side in (12) above, can also be modified in this way. We will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\vec{r}_{C(t1)}}{\Delta t}\right]}$
14. This can be rearranged as:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{M\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}}$
15. But $\mathbf\small{\frac{\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}=\vec{\bar{v}}_{C(\Delta t)}}$
• So (13) becomes: $\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=M\,\,\vec{\bar{v}}_{C(\Delta t)}}$
In general, we can write:
Eq.7.5:
$\mathbf\small{M\,\,\vec{\bar{v}}_{C(\Delta t)}}=m_1 \,\, \vec{\bar{v}}_{1(\Delta t)}+m_2 \,\, \vec{\bar{v}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{\bar{v}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{\bar{v}}_{n(\Delta t)}$
16. In the above discussion, we have considered a time interval of Δt, which is equal to (t₂-t₁)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average velocities
• If this Δt is very small, the division will give instantaneous velocities
• Then the Eq.7.5 will become:
Eq.7.6:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
• All the velocities in this expression, are instantaneous velocities
• It is the velocity at the instant when the reading in the stop-watch is 't'
17. Eq.7.6 can be written as: $\mathbf\small{M\,\,\vec{v}_{C(t)}=\sum {m_i\;\vec{v}_{i(t)}}}$
• Rearranging this, we get:
Eq.7.7: $\mathbf\small{\vec{v}_{C(t)}=\frac{\sum {m_i\;\vec{v}_{i(t)}}}{M}}$
• This is an effective method to find the velocity with which the 'C' moves 

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So we have seen the velocity of the particles in a system. Next we will see acceleration

1. Let a system consist of two particles P and Q
• Let both P and Q be in motion
2. Consider any instant during that motion
• In earlier sections of this chapter, we have seen that, all particles in the system need not be having the same velocity at an instant (Details here)
• So let us assume that, at that instant, P and Q are having different velocities
3. Consider the instant at which the reading in the stop-watch is 't₁'
    ♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't₂'
    ♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t2)}}$
4. If we subtract the 'initial velocity vector' from the 'final position vector', we will get the change in velocity 
• So we can write:
The 'change in velocity' suffered by 'P' during the time interval of (t₂-t₁) 
= $\mathbf\small{\vec{\Delta v}_P=\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}$
5. If we divide the 'change in velocity' by the 'time interval during which the change took place', we will get the average acceleration (Details here)
• Time interval during which the displacement took place = (t₂-t₁) = Δt
• So we can write:
Average acceleration with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{\Delta v}_P}{\Delta t}=\frac{\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{v}_{P(t2)}}{\Delta t}-\frac{\vec{v}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average acceleration with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{Q(\Delta t)}=\frac{\vec{v}_{Q(t2)}}{\Delta t}-\frac{\vec{v}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above accelerations with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}=\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}+m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}$
• From Eq.7.6, this is $\mathbf\small{M\,\,\vec{v}_{C(t2)}}$
• Similarly, the numerator of the second term on the right side will become:
$\mathbf\small{M\,\,\vec{v}_{C(t1)}}$
10. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\left[\frac{M\,\,\vec{v}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\,\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\left[\frac{\vec{v}_{C(t2)}-\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
11. From Newton's second law, We have:
Force = mass × acceleration
• In our present case, the acceleration is average acceleration. So we can write:
Average force = mass × average acceleration
• Thus the result in (10) becomes:
$\mathbf\small{\vec{\bar{F}}_{P(\Delta t)}+\vec{\bar{F}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
• That means:
Average force acting on particle P + Average force acting on particle Q
= [Total mass × Average acceleration of the 'C']
In general we can write:
Eq.7.8:
[Total mass × Average acceleration of the 'C'] 
= $\mathbf\small{\vec{\bar{F}}_{1(\Delta t)}+\vec{\bar{F}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+\vec{\bar{F}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+\vec{\bar{F}}_{n(\Delta t)}}$
11. In the above discussion, we have considered a time interval of Δt, which is equal to (t₂-t₁)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average acceleration
• If this Δt is very small, the division will give instantaneous acceleration
• Then the Eq.7.8 will become:
Eq.7.9:
[Total mass × Instantaneous acceleration of the 'C'] 
= $\mathbf\small{\vec{F}_{1(t)}+\vec{F}_{2(t)}+ \,\,.\,\,.\,\,.+\vec{F}_{i(t)}+\,\,.\,\,.\,\,.+\vec{F}_{n(t)}}$
• All the forces in this expression, are instantaneous forces
• It is the force at the instant when the reading in the stop-watch is 't'
So we can write:
The following two quantities are equal:
(i) The product of the 'total mass' and 'acceleration experienced by 'C''
(ii) The vector sum of all the forces acting on the particles in the system
12. Note:
• $\mathbf\small{\vec{F}_{1(t)}}$ is the force acting on 'particle 1' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 1'. We must calculate the resultant of all those forces
Like wise:
• $\mathbf\small{\vec{F}_{2(t)}}$ is the force acting on 'particle 2' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 2'. We must calculate the resultant of all those forces
• Like wise, we can write for all other particles also
13. This leads us to an interesting result. It can be explained with the help of an example:
(i) A system consists of 2 particles P and Q
• Forces acting on P are: $\mathbf\small{\vec{F}_{P1},\,\vec{F}_{P2}\, \text{and}\,\vec{F}_{P3}}$
• In addition to the above forces, the particle Q also exerts a force on P. This force is $\mathbf\small{\vec{F}_{PQ}}$
    ♦ This force can be considered as an internal force
    ♦ Because it is exerted between particles inside the system
• So the net force acting on P is: $\mathbf\small{\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ}}$
(ii) Forces acting on Q are: $\mathbf\small{\vec{F}_{Q1},\,\vec{F}_{Q2}\, \text{and}\,\vec{F}_{Q3}}$
• In addition to the above forces, the particle P also exerts a force on Q. This force is $\mathbf\small{\vec{F}_{QP}}$
    ♦ This force can be considered as an internal force
    ♦ Because it is exerted between particles inside the system
• So the net force acting on Q is: $\mathbf\small{\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP}}$

(iii) Once we know all the forces acting on each particles, we can apply Eq.7.9

We get:
[m_P+m_Q) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP})}$
(iv) But according to Newton's third law, $\mathbf\small{\vec{F}_{PQ}=-\vec{F}_{QP}}$
• So those two internal forces will cancel each other
• Thus we get:
[m_P+m_Q) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3})}$
■ That means, while applying Eqs.7.8 and 7.9, internal forces in the system have no role to play
14. Consider all the external forces acting on the particles of a system
• Let us denote the vector sum of all the external forces as $\mathbf\small{\vec{F}_{ext}}$
• Also let us denote acceleration of the 'C' as $\mathbf\small{\vec{A}_C}$  
• Then Eq.7.9 becomes:
Eq.7.10: $\mathbf\small{\text{M}\;\vec{A}_C=\vec{F}_{ext}}$
• On the left side, we have: Total mass multiplied by the acceleration experienced by 'C'
• On the right side, we have the vector sum of all external forces acting on the system. That is., the net external force
■ So we can write:
• The net external force will produce the following effect:
Movement of the 'C' with an acceleration, as if, all the mass of the system is concentrated at the 'C' 

• This is the reason why, in the previous chapters, we were able to do problems by considering, wooden blocks, cars, trucks etc., as 'point masses'

16. It may be noted that, if the net force is not acting at 'C', the system will rotate
• We did not consider any rotational motions of objects in the previous chapters
• There was no reason to consider rotation because, we assumed that the net force acts at the 'C'

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A practical example:

1. Consider a projectile shown in fig.7.45(a) below:

Fig.7.45

• We know that, it's path will be a parabola
    ♦ This is indicated by the yellow curve
2. At point 'A', the projectile is intact
• But when it reaches point 'B', it explodes into two pieces 
• Let us call them 'Part P' and 'Part Q'
3. We know that, the only force acting on a projectile is the 'gravitational force'
• This force acts vertically
• There is no horizontal force (Details here)
4. Let us assume that, at the time of launch, P and Q were glued together
• This is shown in fig.b
• The glued combination will have a definite 'C'
    ♦ A force of (m_P × g) acts on P
    ♦ A force of (m_Q × g)  acts on Q
• The glued combination moves as if [(m_P × g)+(m_Q × g)] is acting at 'C' of the combination
5. The combination explodes when it reaches B
(The explosion may be due to the 'ignition of some explosives' kept at the interface between P and Q)
• Beyond B, the parts P and Q move as a system
• Beyond B:
    ♦ the force acting on P is (m_P × g) 
    ♦ the force acting on Q is (m_Q × g)
• So we see that, the forces acting on each part remains the same even after collision
6. P will change course because a force due to explosion is exerted on it
• Q will also change course because a force due to explosion is exerted on it
• But those forces are internal forces and have no contribution on the net force
7. The only external forces are:
• (m_P × g) acting on P
• (m_Q × g)acting on Q
• These forces remain the same before and after the explosion
■ So force acting at 'C' remains the same before and after the explosion
■ Since there is no change in the external force, the  force acting at 'C', does not change
■ So the course of 'C' will not change
■ That means, the location of 'C' will continue as if no explosion have occurred
The following solved example will make this point clear:

Solved example 7.8
A projectile explodes into two pieces P and Q. The explosion occurred at the top most point of it's trajectory. The horizontal distance between the 'launch point' and the 'point of explosion' is x₀. The larger piece Q has 3 times the mass of the smaller piece P. The smaller piece P lands back at the launch point. 
(a) Where does the 'C' of the system land ?
(b) Where does 'Q' land?
Solution:
1. The 'C' of the projectile always lands at a distance of 'R' from the launch point
This is shown in fig.7.46(a) below:

Fig.7.46

• Even if an explosion occur, the 'C' of the fragments will land at the same point
This is shown in fig.b
2. Given that, the explosion occurred at a horizontal distance of x₀ from the launch point
• Also given that, the explosion occurred at the top of the trajectory
3. The top of a parabolic trajectory is at a horizontal distance of $\mathbf\small{\frac{R}{2}}$ from the launch point
• Thus we get: $\mathbf\small{x_0=\frac{R}{2}}$
$\mathbf\small{\Rightarrow R=2x_0}$
• This is the answer for part (a)
4. Now we have the final position of 'C'
• We are given the final position of one of the pieces:
The smaller piece P lands back at the launch point
(The path followed by P is shown as the green dashed curve in fig.c)
5. So we have the position of 'C'
• And we have the position of one of the pieces
6. The position of 'C' is given by: $\mathbf\small{X=\frac{\sum m_i x_i}{M}}$
• To apply this formula, we want a reference frame
• The reference frame shown in fig.7.46 is that of the projectile
• We will use it for 'C' also
• Thus we have:
    ♦ X = R = 2x₀
    ♦ m₁ = m_P
    ♦ m₂ = m_Q = 3m_P
    ♦ x₁ = x_P = 0
    ♦ x₂ = x_Q = ? (The path followed by Q is shown as the red dashed curve in fig.c)
    ♦ M = m_P + mQ = 4m_P
• Substituting these values, we get:
$\mathbf\small{2x_0=\frac{m_P\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_P\times 0+3m_P\,x_Q}{4m_P}=\frac{3}{4}x_Q}$
$\mathbf\small{\Rightarrow x_Q=\frac{8}{3}x_0}$
• This is the answer for part (b)

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So we have seen velocity and force on a system of particles. In the next section, we will see momentum

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Chapter 4.13 - Projectile thrown Horizontally

In the previous section we saw a projectile thrown upwards from a height. In this section, we will see yet another type of Projectile motion.
1. Consider fig.4.34(a) below:

Fig.4.34

• A stone is thrown into the air.
    ♦ It is not thrown straight up.
    ♦ It is not thrown at any angle with the horizontal. 
■ It is thrown in the exact horizontal direction from a height (h) above the ground. 
2. The stop watch is turned on at the instant when the stone is thrown. 
• The position of the stone at that instant is taken as the origin ‘O’. 
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
• The velocity of the stone at O is called initial velocity of the projectile. It is denoted as $\mathbf\small{\vec v_0}$
3. Here, the direction of $\mathbf\small{\vec v_0}$ is exactly horizontal.
• So there will not be a vertical component
• So we get: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{\vec v_0}$    
4. Initially there is horizontal velocity only
• But once the stone leaves O, it will be acted upon by gravity
• So there will be two motions:
    ♦ The horizontal motion with a constant velocity of $\mathbf\small{\vec v_0}$
    ♦ The vertical motion which is under a constant acceleration of g 
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• This vertical component will be affected by the acceleration due to gravity ‘g’
• This is the acceleration vector. We can denote it as (g)$\mathbf\small{\hat j}$
• As a result, magnitude of the vertical component will go on increasing
■ Note that, the initial value of the vertical component is zero. This is because, at O, the stone was given an exact horizontal velocity

6. The horizontal component is responsible for taking the stone ‘horizontally away’ from the origin

• This component is not affected by ‘g’
• So the horizontal component will remain constant during the entire journey.
■ Note that, the air resistance can cause opposition to the projectile motion. But for our present discussion, air resistance is considered to be negligible. So we will not take it into account here.
7. The stone was thrown when the stop watch showed '0' s. What happens to the two velocities when the stop watch shows a reading of 't' seconds?
Ans: The horizontal component will remain the same because, there is no acceleration in the horizontal direction
■ The vertical component will have a larger value because there is positive acceleration (due to gravity) in the vertical direction
• We can find it's exact value at time = 't' s
• For that, we use the familiar equation:  v = v₀ + at
• Thus we can write: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=0+(g)\hat{j}t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(-gt)\hat{j}}$
• The '-' sign is given because, the travel is towards the negative side of the y axis 
• So we can write:
At any time 't', after the beginning of the journey, the magnitude of the vertical component of velocity is given by Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
8. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a higher magnitude as given by Eq.4.20 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a larger magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.34(b). We see the following:
• At time = 't' seconds:
    ♦ The stone has reached P
    ♦ $\mathbf\small{\vec v}$ has a larger length than $\mathbf\small{\vec v_0}$
9. We saw how the 'velocity of the stone' varies during it's travel. Next we will see how 'it's distance from O' varies
■ First we will see the horizontal travel
(i) We have seen that the horizontal velocity remains the same.
• So we can use the familiar 'equation for uniform motion': s = vt
(ii) Thus we get:
Horizontal displacement in time 't' s = $\mathbf\small{\vec{\Delta r_x}=\vec{v_0}\times t}$
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{ | \vec{\Delta r_x} |=\left (  | \vec{v_0}|  \right )t}$
(iv) This magnitude is the distance OP'. But the distance OP' is the x coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left (  | \vec{v_0}|  \right )t}$' from the y axis
• In other words, at any time 't', after the beginning of the journey, the x coordinate of the object is given by Eq.4.21: x = $\mathbf\small{\left (  | \vec{v_0}|  \right )t}$ 
■ Now we will see the vertical travel
(i) The vertical travel is affected by an acceleration 'g'. So we will use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: 
Vertical displacement in time 't' s = $\mathbf\small{0 \times t+\frac{1}{2}g t^2}$
(iv) This magnitude is the distance P'P. But the distance PP' is the y coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\frac{-1}{2}g t^2}$' from the x axis
• In other words, at any time 't', after the beginning of the journey, the y coordinate of the object is given by Eq.4.22: y = $\mathbf\small{\frac{-1}{2}g t^2}$
• The '-' sign is given because, the travel is towards the negative side of the y axis 
10. So we are now able to specify the position of a projectile at any time 't'.
• We are able to do it by using x and y coordinates.
■ If we can eliminate 't' from Eqs.4.21 and 4.22, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.21, we get: $\mathbf\small{t=\frac{x}{|\vec{v_0}|}}$
• We can use this instead of 't' in Eq.4.22. 
• We get Eq.4.23: $\mathbf\small{y=\left [ \frac{-g}{2\left ( | \vec{v_0}  |  \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the square brackets
• 'g' and '2' are constants
• Once the stone is thrown, it's initial velocity canot be changed. So $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the square brackets are constants
• Thus the final result inside those square brackets is a constant. We will denote it as 'a'
(iv) Eq.4.23 becomes: $\mathbf\small{y=ax^2}$
Where $\mathbf\small{a=\left [ \frac{-g}{2\left (  | \vec{v_0}  |  \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
11. Time required for the whole flight (T_f):
■ This is also equal to the time required to reach the ground
(i) Let us consider the vertical motion after O. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h.
(ii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$ 
• In this motion, the initial velocity is zero. It is like the stone just dropped from a height of h
• So we can put v₀ = 0
• We get: $\mathbf\small{h=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow h=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow t=\sqrt{\frac{2h}{g}}}$
• So we get Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
12. Horizontal range of the projectile ($\mathbf\small{|\vec R|}$):
(i) For this we consider the horizontal motion
• The horizontal component of the velocity (which is a constant value) will be effective for the entire time (T_f) of the flight 
(ii) So the horizontal distance = Horizontal velocity × time
= $\mathbf\small{|\vec v_{0x}|\times T_f =|\vec v_{0}|\times T_f = |\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
■ Thus we get Eq.4.25: Range of the projectile $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$

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Now we will see some solved examples
Solved example 4.10
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^-1. Neglecting air resistance, find 
(a) The time taken by the stone to reach the ground, 
(b) Speed with which it hits the ground
(c) Range of the stone (Take g = 9.8 ms^-2)
Solution:
Part (a):
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: T_f = 10 s
Part (b):
1. Magnitude of the horizontal velocity with which the stone hits the ground = $\mathbf\small{|\vec v_x|}$ = 15 ms^-1
2. Magnitude of the vertical velocity with which the stone hits the ground:
• We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
Here t = T_f = 10 s
• Substituting the values, we get: $\mathbf\small{\left | \vec{v_y} \right |}$ = 98 ms^-1. 
3. Speed (magnitude of the resultant velocity) is given by $\mathbf\small{|\vec{v}|=\sqrt{|\vec{v_x}|^2+|\vec{v_y}|^2}}$
• Substituting the values, we get: Speed = 99.14 ms^-1
Part (c):
1. We can use Eq.4.25: $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
2. Substituting the values, we get: $\mathbf\small{|\vec R|}$ = $\mathbf\small{15\times \sqrt{\frac{2 \times 490}{9.8}} }$ = 150 m   
• The path of the stone is shown in fig.4.35 below. It is plotted using Eq.4.23.

Fig.4.35

• We can see that, the coordinates of the point where the stone hits the ground are: (150,-490)

Solved example 4.11
An object is thrown horizontally from the top of a tower. It strikes the ground after 3 seconds at an angle of 45° with the horizontal. Find 
(a) The height of the tower 
(b) The speed with which the object was thrown
[g = 9.8 ms^-2]
Solution:
Part (a):
Given: T_f = 3 s
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: $\mathbf\small{3=\sqrt{\frac{2h}{9.8}}}$
• So h = 44.1 m
Part (b):
Given that, the resultant velocity at the ground makes 45° with the horizontal
1. The angle 45° indicates that magnitudes of both the horizontal and vertical components are equal
• The reason can be given using the following two statements:
(i) tan 45 is always equal to 1
(ii) $\mathbf\small{\tan \theta =\frac{|\vec{v_y}|}{|\vec v_x|}=1}$  Only when $\mathbf\small{|\vec v_y| = |\vec v_x|}$
2. We obtained the height as 44.1 m
• This height was traveled vertically in 3 s
■ What would be the velocity when t = 3 s?
3. We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
• Here t = T_f = 3 s
• So we get: $\mathbf\small{\left | \vec{v_y} \right |=9.8 \times 3 = 29.4\,ms^{-2}}$ 
4. From the result in (1) we get: $\mathbf\small{| \vec{v_x}|}$  = 29.4 ms^-1.
■ Throughout the travel, the horizontal velocity remains the same.
• So we can write: The object was thrown horizontally with a speed of 29.4 ms^-1.

Solved example 4.12
A particle is projected horizontally with a velocity of 20 ms^-1. After what time will the velocity be at an angle of 45° with the horizontal? [g = 10 ms^-2]
Solution:
1. Initially, the particle is projected horizontally
• So initially, it has only the horizontal component
2. But as the travel continues, there will be both horizontal and vertical components.
• They are: $\mathbf\small{\vec{v_x} \: \text{and} \: \vec{v_y}}$
• At any instant after t = 0, the velocity of the particle will be the resultant of those two components
• And that velocity will make an angle with the horizontal
3. In this problem, we are considering the instant at which the resulting velocity makes 45° with the horizontal
• 45° indicates that, $\mathbf\small{|\vec{v_x}|= |\vec{v_y}|}$
4. But $\mathbf\small{|\vec{v_x}|}$ will be always 20 ms^-1.
• So we have to find the instant at which $\mathbf\small{|\vec{v_y}|}$ is also 20 ms^-1
• We can use Eq.4.20: $\mathbf\small{| \vec{v_y} |=gt}$    
• Substituting the values, we get: t = 20/10 = 2 s

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In the next section, we will see circular motion.

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Chapter 4.12 - Projectile motion from a height

In the previous section we saw some properties of Projectile motion. In this section, we will see some solved examples. Solved example 4.9 further below shows the calculations when a projectile is thrown upwards from a height.

Solved example 4.6
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Solution:
• Here, 'elevation' indicates 'initial angle of projection' θ₀.
■ Case 1: Let the angle of projection exceed 45° by x°. 
Then mathematically, the angle of projection θ₀ = (45+x)^o 
■ Case 2: Let the angle of projection fall short of 45° by (the equal amount) x° 
Then mathematically, the angle of projection θ₀ = (45-x)^o
We will consider each case separately. The steps are given below:
Case 1:
1. We can use Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45+x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90+2x) \right )}{g}}$
3. But from math classes, we know that sin (90+x) = cos x
So sin (90+2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
Case 2:
1. We can use Eq.4.19 again: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45-x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90-2x) \right )}{g}}$
3. But from math classes, we know that sin (90-x) = cos x
So sin (90-2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
• This is same as the result in case 1
■ So the two ranges are equal.

Solved example 4.7
A projectile is fired with a speed of 'k' ms^-1. The angle of projection is 57^o. The range obtained is 'R'. Determine the other angle at which the the projectile should be fired with the same speed 'k' to obtain the same range 'R'
Solution:
1. The given angle is 57^o. it exceeds 45 by (57-45) = 12^o
2. The other angle must fall short of 45 by the same amount 12
3. So the other angle = (45-12) = 33^o.

Solved example 4.8
Two objects are projected at angles 45^o and 60^o. The maximum heights reached are the same. What is the ratio of their initial velocities?
Solution:
1. Let
• Magnitude of the initial velocity of object 1 be $\mathbf\small{\left | \vec{{v}_{01}} \right |}$ 
• Magnitude of the initial velocity of object 2 be $\mathbf\small{\left | \vec{{v}_{02}} \right |}$ 
2. Time to reach maximum heights:
• We can use Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
(i) For object 1, we get: $\mathbf\small{t_{m1}=\frac{\left | \vec{v_{01}} \right |\sin 45}{g}}$
$\mathbf\small{\Rightarrow t_{m1}=\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$
(ii) For object 2, we get: $\mathbf\small{t_{m2}=\frac{\left | \vec{v_{02}} \right |\sin 60}{g}}$
$\mathbf\small{\Rightarrow t_{m2}=\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
3. But given that t_m1 = t_m2.
• So we can equate 2(i) and 2 (ii). We get:
$\mathbf\small{\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$ = $\mathbf\small{\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
$\mathbf\small{\Rightarrow \frac{\left | \vec{v_{01}} \right |}{\left | \vec{v_{02}} \right |}=\frac{\sqrt 3}{\sqrt 2}}$

Solved example 4.9
An object is thrown from the top of a building 10 m high. It is thrown upwards with a speed of 25 ms^-1 at an angle of 40^o with the horizontal. 
(a) After what time will it reach the ground? 
(b) What is the distance between the foot of the cliff and the point of impact on the ground?
[g = 9.81 ms^-2]
Solution:
Part (a):
1. As usual, we choose the point of projection as the origin O
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
This is shown in fig.4.33 below:

Fig.4.33

2. We get the same equation Eq.4.15 for the path of the projectile: 
Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0  \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]x^2}$
3. But this time, the path continues to a point Q below the x axis
• This point is at a vertical distance of 10 m below the x axis. This is because, the height of the building is 10 m
4. After being thrown from O, the object meets the x axis again at P
• We can find the time required to reach P from O. 
• We can use Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$  
• Substituting the values, we get: T_f = 3.276 s
5. After passing P, the stone continues the flight for some more time. In the end, it falls back to the ground.
• Let us consider the vertical motion after P. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is 10 m.
■ We can use the familiar equation: $\mathbf\small{s=\left | \vec{v_{0y}} \right |t+\frac{1}{2}\left | \vec{a_y}\right |t^2}$
• Here, s = 10 m. 
• $\mathbf\small{|\vec{v_{0y}}|}$ = the magnitude of the vertical velocity at P 
= magnitude of the vertical velocity at O = 25 sin θ₀ = 16.07 ms^-1
• $\mathbf\small{|\vec{a_{0y}}|}$ = g = 9.81  ms^-2.
6. Substituting the values, we get: 10 = 16.07t + 0.5 × 9.81t²
⇒ 4.905t² + 16.07t - 10 = 0
• Solving this quadratic equation, we get: t = 0.535 s or -3.811 s
• But negative time is not acceptable. So we take t = 0.535 s
7. That is., after passing P, the object travels for 0.535 s more
• So total time of travel = 3.276 + 0.535 = 3.811 s
Part (b):
1. We can find the distance OP using Eq.4.19:
Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
• Substituting the values, we get:
OP = 62.74 m
2. During the last 0.535 s, the object travels horizontally also. The distance covered during this time =
v₀cos θ₀ × t = 25 × cos 40 × 0.535 = 10.245 m
3. So total horizontal distance from O to Q = 62.74 + 10.245 = 72.99 m
• We can see that, the coordinates of Q are (72.99, -10)

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In the next section, we will see another type of projectile motion.

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