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?

Answers (1)

Let us consider the upward direction to be positive

Initial velocity of the ball (u) = 49 m s-1

The speed of the ball will be the same when it reaches the boy's hand's but will be moving in a downward direction. Therefore final velocity (v) = -49 m s-1

Acceleration (a) = -9.8 m s-2

Using the first equation of motion we have

\v=u+at\ t=fracv-ua\ t=frac-49-49-9.8\ t=10 s

In the second case, as the ball has been thrown after the lift has started moving upwards with a constant velocity, the relative velocity of the ball with respect to the boy remains the same and therefore the ball will again take 10 seconds to reach the boy's hands.

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