Question

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

Answer 1

a). As seen from the truck the ball moves vertically upwards corners back. Time taken = time taken by truck to cover $58.8m$.$\therefore time=\frac{s}{v}-\frac{58.8}{14.7}=4sec$ ($V=14.7m/s$ of truck)$u=?v=0,g=-9.8m/s^2$ (going upwards), $t=4/2=2sec$.$v=u+at\Rightarrow 0=u-9.8\times 2\Rightarrow u=19.6m/s$. (vertical upwards velocity).b) From road it seems to be projectile motion.Total time of flight $=4sec$in this time horizontal range covered $58.8m=x$$\therefore x=u\cos \theta t$$\Rightarrow u\cos \theta =14.7$Taking vertical component velocity into consideration$y=\frac{0^2-(19.6{)}^2}{2\times (-9.8)}=19.6m$ [from(a)]$\therefore y=u\sin \theta t-1/2g{t}^2$$\Rightarrow 19.6=u\sin \theta \left(2\right)-1/2\left(9.8\right)2^2\Rightarrow 2u\sin \theta -19.6\times 2$$\Rightarrow u\sin \theta =19.6$$\frac{u\sin \theta }{u\cos \theta }=\tan \theta \Rightarrow \frac{19.6}{14.7}=1.333$$\Rightarrow \theta ={\tan }^{-1}\left(1.333\right)={53}^o$Again $u\cos \theta =14.7$$\Rightarrow u=\frac{14.7}{u\cos {53}^o}=24.42m/s$The speed of ball is $42.24m/s$ at an angles ${53}^o$ with horizontal as seen from the road.