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  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

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 = u cos αt
Here, α  is the angle of projection.
Now,
u cos α = 14.7    ...(i)
Now, take the vertical component of velocity.
Using the equation of motion, we get:

\[v^2 - u^2 = 2\text{ ay }\]

Here, v is the final velocity.
Thus, we get:

\[y = \frac{0^2 - \left( 19 . 6 \right)^2}{2 \times \left( - 9 . 8 \right)}\]
\[ = 19 . 6 \text{ m } \]

Vertical displacement of the ball:

 \[y = u\sin\alpha t - \frac{1}{2}g t^2 \]

\[ \Rightarrow 19 . 6 = u\sin\alpha\left( 2 \right) - \frac{1}{2} \times 9 . 8 \times 2^2 \]

\[ \Rightarrow 2 u \text{ sin } \alpha = 19 . 6 \times 2\]

\[ \Rightarrow u\sin\alpha = 19 . 6 . . . \left(\text{ ii } \right)\]

Dividing (ii) by (i), we get:

\[\frac{u\sin\alpha}{u\cos\alpha} = \frac{19 . 6}{14 . 7}\]
\[ \Rightarrow \tan\alpha = 1 . 333\]
\[\alpha = \tan^{- 1} (1 . 333)\]
\[ \Rightarrow \alpha = 53^\circ\]

From (i), we get:
u cos α = 14.7

\[\Rightarrow u = \frac{14 . 7}{\cos53^\circ} = 24 . 42 \text{ m } /s \approx 25 \text{ m } /s\]

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