Question

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

Answer 1

Given:
Velocity of the truck = 14.7 m/s
Distance covered by the truck when the ball returns again to the truck = 58.8 m
(a) Therefore, we can say that the time taken by the truck to cover 58.8 m distance is equal to the time of the flight of the truck.
Time in which the truck has moved the distance of 58.8 m:
$\mathrm{Time}\left(T\right)=\frac{s}{v}=\frac{58.8}{14.7}=4\mathrm{s}$
We consider the motion of the ball going upwards.
T = 4 s
Time taken to reach the maximum height when the final velocity v = 0:
$t=\frac{T}{2}=\frac{4}{2}=2s$
a = g = −9.8 m/s² (Acceleration due to gravity)
∴ v = u − at
⇒ 0 = u + 9.8 × 2
⇒ u = 19.6 m/s
19.6 m/s is the initial velocity with which the ball is thrown upwards.

(b) From the road, the motion of ball seems to be a projectile motion.
Total time of flight (T) = 4 seconds
Horizontal range covered by the ball in this time, R = 58.8 m
We know:
R = ucosαt
Here, α is the angle of projection.
Now,
ucosα = 14.7 ...(i)
Now, take the vertical component of velocity.
Using the equation of motion, we get:
$v^2-u^2=2ay$
Here, v is the final velocity.
Thus, we get:
$$\begin{gathered} y=\frac{0^2-{\left(19.6\right)}^2}{2\times \left(-9.8\right)} \\ =19.6\mathrm{m} \end{gathered}$$
Vertical displacement of the ball:
$$\begin{gathered} y=u\sin \alpha t-\frac{1}{2}g{t}^2 \\ \Rightarrow 19.6=u\sin \alpha \left(2\right)-\frac{1}{2}\times 9.8\times 2^2 \\ \Rightarrow 2u\mathrm{sin\alpha }=19.6\times 2 \\ \Rightarrow u\sin \alpha =19.6...\left(\mathrm{ii}\right) \end{gathered}$$
Dividing (ii) by (i), we get:
$$\begin{gathered} \frac{u\sin \alpha }{u\cos \alpha }=\frac{19.6}{14.7} \\ \Rightarrow \tan \alpha =1.333 \\ \alpha ={\tan }^{-1}(1.333) \\ \Rightarrow \alpha =53° \end{gathered}$$
From (i), we get:
ucosα = 14.7
$\Rightarrow u=\frac{14.7}{\cos 53°}=24.42\mathrm{m}/\mathrm{s}\approx 25\mathrm{m}/\mathrm{s}$

Therefore, when seen from the road, the speed of the ball is 25 m/s and the angle of projection is 53° with horizontal.