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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8 months ago

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

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According to question, we can write

 10 = 1 2 a t 2 . . . . . . . . . . . . . . . ( 1 ) a n d                        

1 0 + x = 1 2 a ( 2 t ) 2 . . . . . . . . . . . . . . . . ( 2 )  

X = 1 2 A [ ( 2 t ) 2 t 2 ] = 3 ( 1 2 a t 2 ) = 3 0 m  

 

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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