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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?
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?
CBSE | Class 11 | Motion in a Straight Line | NCERT | Physics
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?

Ans: 

According to the question, the initial velocity of the ball (u) is 49 m/s

Case 1 : When the lift is stationary, the boy throws the ball upwards. The positive direction is considered to be vertically upward. The ball’s displacement is zero.

Making use of the equation of motion => s = ut + (1/2)at2

Substituting s = 0, u = 49 m/s and taking a = -g = – 9.8 m/s2 , we get

0 = (49)t + (1/2) (-9.8)t2

Upon re-arranging => t = (49 x 2)/9.8 = 98/9.8

Therefore, t = 10 sec

Case 2 ; In this case, since the lift starts moving with a velocity of 5 m/s, the initial velocity of the ball becomes

Thus, urelative is equal to the velocity of the ballTherefore the time taken in this case is given by the following expression ;t =2urelativegrelative=2×499.8t=10 seconds
This shows that the time taken is same in both the cases.
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