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 ?
Explain me the second part. How can the boy get the ball in the same time wen he's moving up ?
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4 Answers
1it will take 10 seconds even when he is movin up if he throws the ball just after the lift starts moving
1the initial velocity of the ball will now be 49+5=54 m/s in the ground frame
use this to solve the ques you should get the same ans
62The lift moves upwards with a velocity of 5m/s
Suppose that the time for the ball to come back be t.
The initial velocity fo the ball is 49+5=54 m/s in ground frame.
The displacement of the lift in time t= displacemnt of the ball in time t.
Lift moves with constant velocity: Hence displacement=5.t
Ball's displacement will be ut-1/2 g. t2 = 54t - 1/2 g. t2
Both these above quantities are equal...
hence
5.t = 54t - 1/2 g. t2
Thus, 49t=1/2 g. t2
hence, 10t=t2
t=0 or t=10
t=0 refers to the initial moment. and t=10 to the other time when the ball comes back to the lift.
Note that this is the same as in the first case..