Projectile strikes!

Classical Mechanics Level 3

A stone is projected from a horizontal plane. It attains the maximum height H , H, strikes a stationary, smooth wall, and falls on the ground vertically below the point of the maximum height.

Assuming the collision is elastic, what is the height h h of the point on the wall where the ball strikes?

Source: JEE Mock Test

H 4 \dfrac{H}{4} 3 H 8 \dfrac{3H}{8} H 2 \dfrac{H}{2} 3 H 4 \dfrac{3H}{4}

Rishu Jaar by Rishu Jaar , 21, India

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1 solution

Fiki Akbar
Nov 14, 2017

Since the collision is elastic then we can assume there is no collision and particle just continue their motion.

The equation of motion (we only need consider the half of particle motion), x ( t ) = v 0 t y ( t ) = H − 1 2 g t 2 \begin{aligned} x(t) & = & v_{0} t \\ y(t) & = & H - \frac{1}{2}gt^2 \end{aligned}

When particle land in the ground, we have t g = 2 H g t_{g} = \sqrt{\frac{2H}{g}} . From the image above, the horizontal distance between the point of launch and the wall is half of the total horizontal distance. Because the x x is linear respect to t t , then the traveling time until the collision is t c = 1 2 t g = 1 2 2 H g t_{c} = \frac{1}{2}t_{g} = \frac{1}{2} \sqrt{\frac{2H}{g}} .

Substituting into y y equation, we get h = 3 H 4 h = \frac{3H}{4}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir , very nice solution(+1).

Rishu Jaar - 3 years, 6 months ago

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