Question

A ball is projected from a given point with velocity u at some angle with the horizontal and after hitting a vertical wall returns to the same point . Show that the distance of the point from the wall must be less than eu^2 /(e+1)g where e is coeff of restitution

Answer 1

A ball is projected from a given point with velocity u at some angle $\theta$ with horizontal as shown in figure.

so, net displacement along vertical direction , $s_y=0$

$s_y=usin\theta.t-\frac{1}{2}gt^2$

So, t = $\frac{2usin\theta}{g}$

horizontal distance covered by ball in before hitting the wall, $x=ucos\theta. t_1$

or, $t_1=\frac{x}{ucos\theta}$

velocity of ball after hitting the wall , $u_x=eucos\theta$

so, horizontal distance covered by ball after hitting the wall, $x=eucos\theta t_2$

or, $t_2=\frac{x}{eucos\theta}$

here, $t=t_1+t_2$

$\frac{2usin\theta}{g}=\frac{x}{ucos\theta}+\frac{x}{eucos\theta}$

$\frac{2usin\theta cos\theta}{g}=x\frac{(1+e)}{eu}$

$\frac{eu^2}{(1+e)g}sin2\theta=x$

we know, $\frac{eu^2}{(1+e)g}sin2\theta\leq\frac{eu^2}{(1+e)g}$

hence proved that the distance of the point from the wall must be less than eu^2/(e+1)g