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 of projection, find the maximum distance $d$ for which the ball can return to the point of projection.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The vertical force on the ball is only $mg$ throughout its motion but during impact, it experiences a horizontal force from the wall.

$\Rightarrow$ we can use $U_{y}t-\frac{1}{2}g{t}^2$ = $S_y$

Let $t$ = total time of flight $\Rightarrow$ 0 = $usinat-\frac{1}{2}g{t}^2$ $\Rightarrow$ $t$ = $\frac{2usin\alpha }{g}$

Due to impact with the wall at $B$, the normal component (i.e. horizontal component) of velocity is reversed and becomes $e$ times.

$\Rightarrow$ horizontal velocity before impact = $ucos\alpha$ and horizontal velocity after impact = $eucos\alpha$

$\Rightarrow$time taken to reach the wall = $t_1$ = $\frac{d}{ucos\alpha }$

and time taken to come back to $O$ from $B$ = $t_2$ = $\frac{d}{eucos\alpha }{t}_1+t_2$ = $t$

$\Rightarrow$ $\frac{d}{ucos\alpha }+\frac{d}{eucos\alpha }$ = $\frac{2usin\alpha }{g}$ $\Rightarrow$ $u^{2}sin2\alpha$ = $gd[1+\frac{1}{e}]$

$\Rightarrow$ $\frac{gd}{u^2}[1+\frac{1}{e}]\le 1$ $\Rightarrow$ $d\le \frac{e{u}^2}{g(1+e)}$