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Question -

A boy standing on a stationary lift (open from above) throws a ballupwards 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 startsmoving up with a uniform speed of 5 m s-1 and the boy againthrows the ball up with the maximum speed he can, how long does the ball taketo return to his hands?



Answer -

Initial velocity ofthe ball, u = 49 m/s

Case : I

The boy throws theball upwards when the lift is stationary. The vertically upward direction istaken as the positive direction. The displacement of the ball is zero.

Considering theequation of motion

s=ut + (1/2)at2

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

t = (49 x 2)/9.8 =98/9.8 = 10 sec

Case : II

As the lift startsmoving with a speed of 5 m/s, the initial speed of the ball will be 49 m/s + 5m/s = 54 m/s

The displacement ofthe ball will be s =  5t’

Therefore, the timetaken can be calculated using the formula

s = ut + (1/2) at2

5t’ = (54) t’ +(1/2)(-9.8) t’2

t’ = 2(54 – 5)/9.8 =10 sec

The time taken willremain the same in both the cases.

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