A boy standing on a stationary lift (open from above) throws a ball upwards with maximum initial speed he can, equal to 49 m/s. How much time does ball take to return to his hands? If lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

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Solution

Given,
Initial speed of ball = 49
Let after t time body reaches the highest point .
Time of ascent = time of descent
So, total time taken to return the ball his hand = 2t
Use kinematics equation ,
V = u + at
0 = 49 - 9.8t
t = 49/9.8 = 5 sec
Hence, time taken to return his hand = 2 x 5 = 10sec

When lift is moving upward with uniform speed . Hence acceleration of lift will be zero. Then relative velocity of ball with respect to boy as same as velocity of the ball .
Hence time taken by ball to return his hand = 10sec

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