Question

A stone is projected from a horizontal plane. It attains maximum height H and strikes a stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic. The height of the point on the wall where ball will strike is

• A. $\frac{2H}{3}$
• B. $\frac{3H}{4}$
• C. $\frac{H}{4}$
• D. $\frac{H}{2}$

Answer 1

The correct option is B

$\frac{3H}{4}$

$usin\theta =\sqrt{2gH}$

$T=\frac{2usin\theta }{g}=\frac{2\sqrt{2gH}}{g}$

The horizontal distance covered

$T=t_1+t_2$

$\frac{2usin\theta }{g}=\frac{(\frac{R}{2}+x)}{ucos\theta }+\frac{x}{ucos\theta }\Rightarrow R=\frac{R}{2}+x+x\Rightarrow x=\frac{R}{4}$

$t_1=\frac{3R}{4ucos\theta }=\frac{3}{4}2\sqrt{2gH}$

$=\frac{3}{2}\sqrt{\frac{2H}{g}}$

$=3\sqrt{\frac{H}{2g}}$

$h=usin\theta \times t_1-\frac{1}{2}g{t}_1^2$

$=\sqrt{2gH}\times 3\sqrt{\frac{H}{2g}}-\frac{1}{2}g\times \frac{9H}{2g}$

$=3H-\frac{9H}{4}=\frac{3H}{4}$