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Motion

Section: Physics

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Question : 7 of 13

Marks: +1, -0

A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meters, is

[JEE Adv 2018 P2]

Your Answer:

Solution:

H_1=\frac{u^2\sin^{2}45}{2g}=120\;\;\;\;\Rightarrow \frac{u^2}{4g}=120

.....(i)

When half of kinetic energy is lost

V=\frac{u}{\sqrt{2}}

H_2=\frac{\left(u/\sqrt{2}\right)^2\sin^{2}30}{2g}=\frac{u^2}{16g}

........(ii) From (i) and (ii)

H_2=30.00

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