A juggler projects a ball upward in a room of height H = 20 m . Ball's projection speed was such that it can just reach the ceiling. Now the ball is projected with two times of the previous veloc

Question

A juggler projects a ball upward in a room of height $H = 20 m$ . Ball's projection speed was such that it can just reach the ceiling. Now the ball is projected with two times of the previous velocity. After what time interval (in second) does this ball return to hand of juggler. Assume collision with roof results in reversal in direction of velocity with no change in magnitude.

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Answer:

A

Solution:

$v_{0} = 2 gH$ $H^{'} = \frac{( 2 v _{0} ) ^{2}}{2 g} = 4 H ∴ t = 2 [\frac{2 ( 4 H )}{g} - \frac{2 ( 3 H )}{g}] = 8 - 43$

TestHub · Accepted Answer

v 0 ​ = 2 gH ​ ​ H ′ = 2 g ( 2 v 0 ​ ) 2 ​ = 4 H ​ ∴ t = 2 [ g 2 ( 4 H ) ​ ​ − g 2 ( 3 H ) ​ ​ ] = 8 − 4 3 ​ ​

Physics - Kinematics Question with Solution | TestHub

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