Q 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?

Answers (1)
S Sayak

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=\frac{v-u}{a}\\ 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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