A stone is projected from a horizontal plane. It attains maximum height 'H' & strikes a stationary smooth wall & 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:

Engineering

Physics

Acceleration

Collision

Projectile Motion

Question

A stone is projected from a horizontal plane. It attains maximum height 'H' & strikes a stationary smooth wall & 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:

3H/4

H/4

None

H/2

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$t = \frac{3 R}{4 vcosθ} = \frac{3 \times 2 v^{2} sinθ \cdot cosθ}{4 vcosθg} = \frac{3 vsinθ}{2 g}$

h = u_yt – gt² = v sin θ . $(\frac{3 v \sin θ}{2 g}) - \frac{1}{2} g {(\frac{3 v \sin θ}{2 g})}^{2}$

$= \frac{3 v^{2} \sin^{2} θ}{2 g} - \frac{9 v^{2} \sin^{2} θ}{8 g}$

$= \frac{v^{2} \sin^{2} θ}{2 g} (3 - \frac{9}{4}) = \frac{3 H}{4} (∵ H = \frac{v^{2} \sin^{2} θ}{2 g})$ .