Question

A ball is thrown horizontally from the top of a tower of unknown height. The ball strikes a vertical wall whose plane is normal to the plane of motion of the ball. The collision is elastic and the ball falls on the ground exactly at the midpoint between the tower and the wall. The ball strikes the ground at an angle ${30}^{\circ }$ with the horizontal. Find the height of the tower (in meters).

Answer 1

AP is the path of the projectile in the absence of the wall, which is half of a normal projectile with range $R=2OP$ and angle of projection $\theta ={30}^{\circ }$ and maximum height $H$.

Because collision with the wall is elastic, horizontal velocity of the ball is reversed.

$\therefore B{P}^{{}^{\prime }}=O{P}^{{}^{\prime }}=\frac{1}{2}OB=BP$
$\Rightarrow OP=R=\frac{3}{2}OB$
$\Rightarrow R=12\sqrt{3}\text{m}$

$R=\frac{u^2\sin 2\theta }{g}$,
$\therefore H=\frac{u^2{\sin }^2\theta }{2g}$
$=\frac{1}{2}\left(\frac{{\sin }^2\theta }{\sin 2\theta }\right)R=6\text{m}$