A person is standing on a truck moving with a constant velocity of $14.7 \mathrm{~m} / \mathrm{s}$ on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved $58.8 \mathrm{~m}$. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.
Time taken by ball=Time taken by truck to cover $58.8 \mathrm{~m}$
Time= $\frac{\underline{\text { distance }}}{\text { speed }}=\frac{58.8}{14.7}=4 \mathrm{sec}$
(a)
For Truck, ball will seems to travel in vertical direction
Time to reach $\mathrm{H}_{\max }$ by ball= $=\frac{\frac{4}{2}}{2}=2 \mathrm{sec}$
$v=0 ; a=-g ; t=2 \sec$
$v=u+a t$
$0=u-g \times 2$
$0=\mathrm{u}-\mathrm{g} \times 2$
$\mathrm{u}=19.6 \mathrm{~m} / \mathrm{s}$ (upward direction)
(b)
From road, motion of ball will be projective
Now, usin $\theta=19.6$
$U \cos \theta=14.7$
Squaring and adding
$u^{2}=(19.6)^{2}+(14.7)^{2}$
$u=25 \mathrm{~m} / \mathrm{s}$
$\operatorname{Tan} \theta=\frac{19.6}{14.7}$
$\theta=53^{\circ}$
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