1it will take 10 seconds even when he is movin up if he throws the ball just after the lift starts moving
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
11the 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..