Question

A ball is projected with a velocity $u$ at an elevation from a point distance $d$ from a smooth vertical wall in a plane perpendicular to it. After rebounding from the wall, it returns to the point projection, proves that $u^{2}sin2\alpha =gf(1+1/e)$. Hence find the maximum distance $d$ for which the ball can return to the point of projection.

Answer 1

The only vertical force on the ball is $mg$ throughout its motion because during impact it experiences a horizontal force from the wall. We can use
$u_{y}1-\frac{1}{2}g{r}^2={\nu }_y$
Let $t$ be the total time of flight.
$\therefore 0=usin\alpha t-\frac{1}{2}g{t}^2$
$⟹t=\frac{2usin\alpha }{g}$
Due to impact with the wall at $B$, the normal component (i.e. horizontal component) of velocity is reversed and because $e$ times.
Horizontal velocity before impact $=ucos\alpha$
and horizontal velocity after impact $=eucos\alpha$
Time taken to reach the wall, $t_1=\frac{d}{\left(ucos\alpha \right)}$
and time taken to come back to $O$ from $B$,
$t_2=\frac{d}{\left(eucos\alpha \right)}$we have $t_1+t_2=t$
$⟹\frac{d}{ucos\alpha }+\frac{d}{eucos\alpha }=\frac{2usin\alpha }{g}⟹u^{2}sin2\alpha =gd[1+\frac{1}{e}]$
As $sin2\alpha \le 1$,
$\frac{gd}{u^2}[1+\frac{1}{e}]⟹\le 1⟹\le \frac{e{u}^2}{g(1+e)}$