Chapter 5.3 - Relation between Velocity and Force

In the previous section we obtained the relation between force and mass. We saw that force is directly proportional to mass. In this section, we will see the relation between force and velocity.

■ We obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and v by:
• Changing v
• Keeping m and t constant
We will do a series of experiments here also:
Experiment 7:
1. Consider a car on a level road. See fig.5.6(a) below:

Fig.5.6

• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F₇ be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The car will not move at the same instant when F₁ is applied. It will take some time to start moving
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F₇ is applied
(ii) The instant when the car attains a velocity of say 2 ms^-1 
(we can fix any convenient value for the velocity. 2 ms^-1 is only an example)
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the speedometer reading reaches 2 ms^-1
• Let t₂ = t
• Then Δt = (t₂ - t₁) = (t-0) = t s
7. So we can write:
• A force F₇ is required to push a car from rest and to move it with a velocity of 2 ms^-1

• The time required for this velocity change (from zero to 2 ms^-1) is t s

Experiment 8:
1. Consider the same car on the same road. See fig.5.6(b) above
• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The car will not move at the same instant when F is applied. It will take some time to start moving
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F is applied
(ii) The instant when the car attains a higher velocity than in experiment 7. Say 3 ms^-1 
(we can fix any convenient value greater than 2 which we used in the previous experiment. 3 ms^-1 is only an example)
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the speedometer reading is 3 ms^-1
• Let t₂ = t
• Then Δt = (t₂ - t₁) = (t-0) = t s
7. But now there is a problem.
• We want both mass and time to be the same as in experiment 7. 
• That is., we want to attain the velocity 3 ms^-1 in the same interval 't' obtained in experiment 7
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the car to rest and start pushing it again 
8. We do the trials until the following two conditions are satisfied
(i) The car attains a velocity of 3 ms^-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 7  
• The force F which satisfies the both two conditions can be noted down as F₈
■ We can write:
• A force F₈ is required to push a car from rest, to move it with a velocity of 3 ms^-1.
• The time required for this velocity change (from zero to 3 ms^-1) is the same t s as in experiment 7

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• The experiment 8 is over

Now we make a comparison between the results of the two experiments 

■ We will find that F₇ < F₈.
The following points may be noted:
(i) In experiment 7, v₁ = 0 and v₂ = 2
• So change in velocity achieved = (v₂ - v₁) = (2-0) = 2 ms^-1
(ii) In experiment 8, v₁ = 0 and v₂ = 3
• So change in velocity achieved = (v₂ - v₁) = (3-0) = 3 ms^-1
(iii) The 'change in velocity' is obviously greater in experiment 8
■ We can write: When 'change in velocity' increases, greater force is required. 
Let us do another set of two experiments to confirm this:

Experiment 9:
1. Consider a 'trolley carrying a mass' on a level road. See fig.4.7(a) below:

Fig.5.7

• It is in a state of uniform motion towards the right.
• It's velocity is 2 ms^-1. 
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F₉ be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The trolley will not stop at the same instant when F₉ is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F₉ is applied
(ii) The instant when the trolley attains a velocity of 0 ms^-1 
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the trolley comes to rest
• Let t₂ = t
• Then Δt = (t₂ - t₁) = (t-0) = t s
7. So we can write:
• A force F₉ is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms^-1 to zero) is t s

Experiment 10:
• We repeat the experiment with the same trolley used in the previous experiment 9. See fig.4.7(b) 
• The mass contained inside it should not change. 
• But the velocity should be different. Let the trolley be moving with a uniform velocity of 3 ms^-1.
• The following points should be noted:
    ♦ We want to bring the trolley to rest 
    ♦ We want to achieve this 'velocity change' (from 3 to zero) within the same duration 't' that we obtained in experiment 9

• We will write the steps:
1. Consider the 'trolley with the same mass' on the same floor as in experiment 9
• It is in a state of uniform motion.
• It's velocity is 3 ms^-1. 
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The trolley will not stop at the same instant when F is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F is applied
(ii) The instant when the trolley attains a velocity of 0 ms^-1 
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the trolley comes to rest
• Let t₂ = t
• Then Δt = (t₂ - t₁) = (t-0) = t s
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 9 
• That is., we want to attain the velocity 0 ms^-1 in the same interval 't' obtained in experiment 9
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 3 ms^-1' and try to stop it again 
8. We do the trials until the following two conditions are satisfied
(i) The trolley attains a velocity of 0 ms^-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 9  
• The force F which satisfies the both two conditions can be noted down as F₁₀
■ We can write:
• A force F₁₀ is required to bring the trolley to rest.
• The time required for this velocity change (from 3 ms^-1 to zero) is the same t s obtained in experiment 9

---

• The experiment 10 is over

Now we make a comparison between the results of the two experiments 

■ We will find that F₉ < F₁₀
The following points may be noted:
(i) In experiment 9, v₁ = 2 and v₂ = 0
• So change in velocity achieved = (v₂ - v₁) = (0-2) = -2 ms^-1.
(ii) In experiment 8, v₁ = 3 and v₂ = 0
• So change in velocity achieved = (v₂ - v₁) = (0-3) = -3 ms^-1.
(iii) The magnitude of the 'change in velocity' is obviously greater in experiment 10
■ We can write: When 'change in velocity' increases, greater force is required. 
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:

Experiment 11:
1. Drop a small stone from the top of a building 
2. Let a person standing at the foot of the building catch it 
3. The following points should be noted while catching:
    ♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
    ♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F₁₁ 
Experiment 12:
1. Drop the same stone which was used in experiment 11
• But this time, the drop must be made from a taller building than in experiment 11
2. Let the person standing at the foot of the building catch it 
3. The following points should be noted this time also:
    ♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
    ♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F₁₂.

---

• Let us make a comparison between the results of experiments 11 and 12

■ We will find that F₁₁ < F₁₂
• The following points may be noted:
(i) The stone dropped from a greater height would have attained a greater velocity (v₂)
    ♦ The initial velocity (v₁) is zero in both cases
    ♦ So the change in velocity (v₂-v₁) will be more in experiment 12
(ii) The same stone was dropped in both the experiments
    ♦ So the mass remains the same
(iii) The person does not lower his hands
    ♦ So the 'time of application of the force' is same in both cases

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■ So we completed three sets of experiments. We can now write with confidence:

• When 'change in velocity (v₂-v₁)' increases, greater force is required 
• When 'change in velocity (v₂-v₁)' decreases, lesser force is required
In other words:
■ Force is directly proportional to 'change in velocity (v₂-v₁)'
• Symbolically, we write this as: F∝ (v₂-v₁)

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■ Thus we found the relation between 'change in velocity' and force. 

• We obtained it by changing the 'velocity' while keeping mass and time constant
• In the next section we will obtain the relation between time and force
• For that, we will change the time while keeping mass and velocity constant

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Chapter 5.1 - Newton's First Law

In the previous section we saw the 'property of inertia' which was put forward by Galileo. In this section, we will see how Sir Isaac Newton based his studies on that concept.

• The three laws put forward by Newton laid the foundation of Mechanics. 
    ♦ The law of inertia put forward by Galileo was the starting point of newton's works.
■ Newton's first law states that:
Every body continues to be in it's state of rest or of uniform motion in a straight line unless compelled by an external force to act otherwise.
• The law is written in one sentence. But it contains many points. 
• We can elaborate it by the following 4 steps: 
1. A body can be in any one of the following two states:
    ♦ The body can be at the state of rest
    ♦ The body can be at the state of uniform motion in a straight line
2. Which ever be the state, it will continue to be in that state
3. If a 'change in it's state' is required, we must apply an external force on that body 
4. As we have seen in the previous section,
    ♦ If the external forces acting on the body cancel each other, the 'change in state' will not be achieved' 
    ♦ To achieve the 'change in state', there must be a net external force.

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■ Now, if a body is at rest, it means that, it has no acceleration
■ If the body is in uniform motion, then also it means that, it has no acceleration
We can all the above information symbolically:
    ♦ No net force ⇒ The body remains at rest ⇒ No acceleration
    ♦ No net force ⇒ The body remains at uniform motion ⇒ No acceleration
[The symbol '⇒' stands for 'implies']
• We can connect the first and the third. We get:
    ♦ No net force ⇒ No acceleration
■ That is., if no net force act on a body, that body will have zero acceleration
• We can write the converse: 
If we want a body to have no acceleration, we must not apply a net force on it

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From the above discussion, the following points become clear:
■ If we apply a net force on a body, it will experience acceleration 
We can write the converse of this also:
■ If a body experiences an acceleration, then a net force is acting on the body
Let us see some practical applications of the above information:
Example 1:
1. A child pulls a non-electric toy car by a string 
See fig.5.3(a) below:

Fig.5.3

• Let the toy car move with uniform velocity
• Then the car experience zero acceleration
2. Based on Newton's First law, we can write:
■ No net force is acting on the car
3. But we must analyse the situation and write the reason for 'no net force and hence no acceleration '
• We see that the child is applying a force through the string. Even then we say that, net force is zero. 
• So what happened to the force applied through the string?
• The answer is that, the 'force applied through the string' is canceled by 'another force in the opposite direction'.
4. This 'another force' is the frictional force which is acting at the interface between the tyres and the floor
• The child tries to pull the car forward
• The frictional force oppose this forward motion
5. The two forces obviously have opposite directions
• So if the two forces are equal in magnitude, they will cancel each other
• If the child's pull is greater in magnitude, then the car will move with acceleration.
■ We can give the inference by writing the three statements below:
1. The car is observed to be moving with uniform velocity. 
2. So the net external force on it must be zero
3. By Newton's first law, we conclude that, the following two  forces are equal in magnitude but opposite in direction:
(a) Force applied by the child
(b) Frictional force at the interface between the wheels and the floor

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■ We must not write the inference in this way:
1. The following two  forces are equal in magnitude but opposite in direction:
(a) Force applied by the child
(b) Frictional force at the interface between the wheels and the floor
2. The two forces cancel each other 
3. So we observe the car to be moving with uniform velocity
■ Why are we not able to write the inference by the above three statements?
Ans: When beginning to solve a problem, we may not be knowing all of the following items:
(i) The forces which are acting
(ii) The magnitudes of the forces
(iii) The directions of the forces
• It is by the 'application of Newton's first law', that we make an inference:
The frictional force is equal in magnitude but opposite in direction to the applied force

---

Example 2:
1. A book rests on a table. See fig.5.3(b) above
• It has no motion at all
• Then the book has zero acceleration
2. Based on Newton's First law, we can write:
■ No net force is acting on the book
3. But we must analyse the situation and write the reason for 'no acceleration and hence no net force'
• In example 1 above, we see a 'visible force' which is applied through a sting. 
• We asked: What happened to that force?
• And we found the answer
• But here, we see no 'visible force'
• We are inclined to conclude that:
The case of 'no net force' need not be considered here. Because no force is acting
4. But on all objects on earth, the gravitational force is acting. 
• Because of this force, the book will be pulled down wards. 
• The magnitude of this force is equal to 'W', the weight of the book
5. So now we have to consider 'net force'
• We have a force 'W' acting on the book
• Yet we see no acceleration
• What happened to 'W'?
• The answer is that, 'W' is cancelled by 'another force in the opposite direction'.
6. This 'another force' is the reaction 'R' exerted by the table
• It is easy to see that, if the table is not present to provide 'R', the book will fall 
• The 'W' tries to pull the book downwards
• The 'R' opposes this downward motion
■ We can give the inference by writing the three statements below:
1. The book is observed to be at rest. 
2. So the net external force on it must be zero
3. By Newton's first law, we conclude that, the following two forces are equal in magnitude but opposite in direction:
(a) The weight W of the book
(b) The reaction R exerted by the table

---

■ We must not write the inference in this way:
1. The following two  forces are equal in magnitude but opposite in direction:
(a) The weight W of the book
(b) The reaction R exerted by the table
2. The two forces cancel each other.
3. So we observe the book to be at rest
■ Why are we not able to write the inference by the above three statements?
Ans: When beginning to solve a problem, we may not be knowing all of the following items:
(i) The forces which are acting
(ii) The magnitudes of the forces
(iii) The directions of the forces
• It is by the 'application of Newton's first law', that we make an inference:
The Reaction R is equal in magnitude but opposite in direction to the weight W

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■ We see that the property of inertia put forward by Galileo is contained in Newton's first law
• We experience inertia in many day to day situations
Let us see an example. We will write it in steps: 
1. Consider a person standing inside a bus
• Initially, the bus is at rest
• When the bus starts to move forward, the person tends to fall backwards
2. This can be explained as follows: 
• The feet are in contact with the floor. 
• When the floor moves forward the feet (due to inertia), would want to stay at rest.
• The feet would not want to move with the floor. 
3. But the friction (between the feet and the floor) will not allow the feet to stay at rest. 
• It will carry the feet forward. 
4. This motion of the feet should carry the entire body forward. 
• But the human body is somewhat flexible. It is not a rigid object. 
• So the upper parts will not experience the same motion of the feet.
5. The upper parts like to stay at rest, and somewhat succeeds in doing so. 
• But the feet is not present straight below to carry the upper part. It has moved forward. 
• So the person falls back
6. The opposite happens when the bus stops.
• The feet comes to stop due to friction. 
• The upper parts tend to continue being in the state of motion. 
• But the feet has stopped moving. There is no feet straight below to carry the upper part. 
• So the person falls forward

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In the next section, we will see how Newton's second law.

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Chapter 4.10 - Projectile Motion

In the previous section we saw 2-dimensional motion under constant acceleration. In this section, we will see Projectile motion.
1. Consider fig.4.29(a) below:
 

Fig.4.29

• A stone is thrown into the air. 
• It is not thrown straight up. But at an angle (less than 90^o) with the horizontal. 
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. Once the stone is thrown, it is on it’s own. That is., once it is thrown, no propelling force acts on it.
(In a rocket, it’s engine produces exhaust gas which propels it forward. Such a motion is not considered as projectile motion)
4. We see that the stone is thrown at an angle.
• So the initial velocity will have a vertical component $\mathbf\small{\vec v_{0y}}$ and a horizontal component $\mathbf\small{\vec v_{0x}}$
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• But 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 velocity component will go on decreasing.   
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 of 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. At O, let $\mathbf\small{\theta_0}$ be the angle made by $\mathbf\small{\vec v_0}$ with the horizontal. Then at O:
• The horizontal component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ 
• The vertical component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0y}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\sin \theta _0)}\hat{j}$
8. The stone was thrown when the stop watch showed '0' s. What happens to these components 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 smaller value because there is negative 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: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0)\hat{j}-(g)\hat{j}t}$    
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0-gt)\hat{j}}$
• 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.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt)}$
9. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a lower magnitude as given by Eq.4.12 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a smaller magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.29(b). We see the following:
• At time = 't' seconds:
    ♦ The stone has reached P
    ♦ $\mathbf\small{\vec v}$ has a smaller length than $\mathbf\small{\vec v_0}$
    ♦ $\mathbf\small{\theta}$ is different from $\mathbf\small{\theta_0}$
10. 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 = $\mathbf\small{\vec{\Delta r_x}=(\left | \vec{v_0} \right |\cos\theta_0 )\hat{i}\times t}$
$\mathbf\small{\Rightarrow \vec{\Delta r_x}=[(\left | \vec{v_0} \right |\cos\theta_0 )t]\hat{i}}$  
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{\left | \vec{\Delta r_x} \right |=\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$
(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 '$\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$' 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.4.13: x = $\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$ 
■ 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 = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
(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 '$\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$' 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.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
11. 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.4.13 and 4.14, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.13, we get: $\mathbf\small{t=\frac{x}{\left |\vec{v_0}  \right |\cos \theta _0}}$
• We can use this instead of 't' in Eq.4.14. 
• We get Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0  \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the first pair of square brackets
• $\mathbf\small{\theta_0}$ is the initial angle with which the stone is thrown at the beginning
• Once the stone is thrown at a particular initial angle, it will not be altered
• That is., $\mathbf\small{\theta_0}$ is a constant. 
• So $\mathbf\small{\tan \theta_0}$ is a constant. We will denote it as 'a'   
(iii) Consider the quantities inside the second pair of square brackets
    ♦ 'g' and '2' are constants
    ♦ $\mathbf\small{\theta_0}$ is a constant as we saw above
• Now  $\mathbf\small{\left | \vec{v_0} \right |}$ remains
    ♦ Once the stone is thrown at a particular initial velocity, it will not be altered
    ♦ That is., $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the second pair are constants
• So the final result inside that second pair is a constant. We will denote it as 'b'
(iv) Eq.4.15 becomes: $\mathbf\small{y=ax+bx^2}$
Where $\mathbf\small{a=\left [ \tan\theta_0  \right ]\: \: \text{and}\; \; b=-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax+bx^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
12. Time required to reach the maximum height:
• Consider the path of the projectile shown in fig.c
• We see a peak point M. After M, we see no further upward motion
• That means, at this peak point M, the magnitude of the vertical component is zero
(i) Let $\mathbf\small{t_m}$ be the time required to reach M
(ii) Consider the vertical component of the velocity. We can use Eq.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$
(iii) Substituting the known values, we get: $\mathbf\small{0=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$   
■ From this we get Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
13. Maximum height reached by the stone (h_m):
(i) For this, we need the height of M from the x axis
(ii) Let us use Eq.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
• In this equation, if we put t = t_m, we will get the vertical distance traveled during 't_m'
(iii) But the 'vertical distance traveled during t_m' is the height of M. So we get:
$\mathbf\small{y=h_m=\left ( \left |\vec v_0 \right | \sin \theta_0 \right )\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )-\frac{g}{2}\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )^2}$
• From this, we get Eq.4.17: $\mathbf\small{h_m=\frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}}$
14. Time required for the whole flight (T_f):
(i) After M, 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 M. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h_m.
(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_m
• So we can put v₀ = 0
• We get: $\mathbf\small{h_m=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{g^2}=t^2}$
$\mathbf\small{\Rightarrow t=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
• So total time of flight = T_f = (t_m+t) = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}+\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$ 
• So we get Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$
■ Note:
• From Eq.4.16, we have: Time required for the upward travel from O to M = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• In the above step (13), we have: Time required for the downward travel from M to the ground = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• So the times for upward travel and downward travel are the same
15. Horizontal range of a 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 component of velocity × time
= $\mathbf\small{\vec v_{0x}}$ × $\mathbf\small{T_f}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ × $\mathbf\small{\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$ =$\mathbf\small{\frac{\left ( \left | \vec{v_0} \right |^2 2 \sin \theta_0 \cos \theta_0 \right )\hat{i}}{g}}$
• But from math classes, we have: $\mathbf\small{2 \sin \theta_0 \cos \theta_0}$ = $\mathbf\small{\sin 2\theta_0}$
(iii) So we can write: $\mathbf\small{\vec{R}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )\hat{i}}{g}}$
(iv) Magnitude of $\mathbf\small{\vec R}$ is the actual distance
■ Thus we get Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
16. Maximum possible range for a given velocity:
• Suppose that a machine can throw an object only at a certain 'fixed speed' $\mathbf\small{\left |\vec v_0 \right |}$
    ♦ But the angle of projection can be changed to any value.
[That is., magnitude of $\mathbf\small{\vec v_0}$ is fixed. But the direction can change]
■ Then what angle would we choose to obtain 'maximum range'?
Solution:
1. We have:
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. In our present case, $\mathbf\small{\left |\vec v_0 \right |}$ is a constant
• So the only variable is sin 2θ₀.
3. That means, for maximum range, sin 2θ₀ must be maximum
• The maximum value possible for sin 2θ₀ is '1'. 
4. This '1' is obtained when '2θ₀' is 90^o.
• So θ₀ must be 45^o.
■ We can write:
The maximum range is obtained when the angle of projection θ₀ is 45^o.

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In the next section, we will apply the above equations to an actual projectile.

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