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:
• A stone is thrown into the air.
• It is not thrown straight up. But at an angle (less than 90o) 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 (hm):
(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 = tm, we will get the vertical distance traveled during 'tm'
(iii) But the 'vertical distance traveled during tm' 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 (Tf):
(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 hm.
(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 hm
• 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 = Tf = (tm+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 (Tf) 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 90o.
• So θ0 must be 45o.
■ We can write:
The maximum range is obtained when the angle of projection θ0 is 45o.
1. Consider fig.4.29(a) below:
 |
| Fig.4.29 |
• It is not thrown straight up. But at an angle (less than 90o) 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 (hm):
(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 = tm, we will get the vertical distance traveled during 'tm'
(iii) But the 'vertical distance traveled during tm' 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 (Tf):
(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 hm.
(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 hm
• 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 = Tf = (tm+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 (Tf) 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 90o.
• So θ0 must be 45o.
■ We can write:
The maximum range is obtained when the angle of projection θ0 is 45o.
In the next section, we will apply the above equations to an actual projectile.
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