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?
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?
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1 Answer
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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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Please find the solution below:
[h] = ML2T-1
[E] = ML2T-2
[V] = ML2T-2C-1
[P] = MLT-1
According to question, we can write
10 =
Average speed
(d) Initial velocity
Final velocity
Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
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