3.24 A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s–1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s-1 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?

0 93 Views | Posted 5 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    5 months ago

    3.24 The initial velocity of the ball, u = 49m/s

    First Case: When the ball returns to his hands, total displacement = 0

    From the relation s = ut + 0.5at2, we get 0 = 49t + 0.5 (-9.81) t2

    4.905 = 49t      Hence t = 10s

    Second Case:

    As the lift started moving up with a speed of 5 m/s, the initial velocity of the ball = 49 + 5 m/s = 54 m/s

    If t' is time for the ball to return to his hand, the displacement of the ball will be = 5t'

    From the relation        s = ut + 0.5 x at2, we get

    5t' = 54t' + 0.5 * (-9.8) t'2

    49t' = 4.9 t'2

    t' = 10 s      

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