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. Assuming the collision to be elastic the height of the point on the wall where ball will strike is

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. Assuming the collision to be elastic the height of the point on the wall where ball will strike is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0d8f7745c4fa26e742c813e65c3e3363-.png" /> (A) (H/2) (B) (H/4) (C) (3 H/4) (D) None of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bfdefa018459dca0ef09241753eaaf08-.png" /> H=(1/2) g (2t)2=2gt2 dots (1) h=H-(1/2) gt2 dots (2) Solving eqns (1) and (2), we get h=H -(H/4)=(3H/4)

Q. 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. Assuming the collision to be elastic the height of the point on the wall where ball will strike is

$\frac{H}{2}$

$\frac{H}{4}$

$\frac{3 H}{4}$

None of these

$H = \frac{1}{2} g (2 t)^{2} = 2 g t^{2} \dots (1)$
$h = H - \frac{1}{2} g t^{2} \dots (2)$
Solving eqns $(1)$ and $(2)$ , we get
$h = H - \frac{H}{4} = \frac{3 H}{4}$