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) Aa seen from the road.

Answer 1

A seen from the truck the ball moves vertically upward comes back.

Time taken = time taken by truck to cover 58.8 m

$\therefore Time=\frac{s}{v}=\frac{58.8}{14.7}$ = 4 sec

Hence v = 14.7 m/s for truck

u = ?, v = 0 = -9.8 $m/s^2$

(going upward)

t = \frac{1}{2} = 2 sec

$\Rightarrow 0=u+9.8\times 2$

$\Rightarrow$= 19.6 m/s

(vetical upward velocity)

(b) From road, it seems to be projectile motion.

Total time of flight = 4 sec

Horizontal range covered in this time

= 58.8 m = X

$\therefore$ x = u cos $\theta$

$\Rightarrow ucos\theta t=14.7$ ....(1)

Taking vertical componene of velocity into consideration

y = $\frac{0^2-\left({19.6}^2\right)}{2\times (-9.8)}$

= 19.6 m [from (a)]

$\therefore y=usin\theta t-\frac{1}{2}g{t}^2$

$\Rightarrow 19.6=usin\theta \left(2\right)-\frac{1}{2}\left(9.8\right)2^2$

$\Rightarrow 2usin\theta =19.6\times 2$

$\Rightarrow usin\theta$ = 19.6 ...(2)

Dividing equation (2) by equation (1), we get)

$\frac{usin\theta }{ucos\theta }=tan\theta =\frac{19.6}{14.7}$ = 1.333

$\Rightarrow \theta =ta{n}^{-1}\left(1.333\right)={53}^0$

Again, u cos $\theta =14.7$

$\to u=\frac{14.7}{cos{5.}^3}=24.42m/s$

The speed of ball is 24.42 m/s at an angle ${53}^0$ with horizontal as seen from the road.