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A very broad elevator is going up vertically with a constant acceleration of 2 m/s2 . At the instant when it velocity is 4 m/s a ball is projected from the floor of the lift with a speed of 4 m/s relative to the floor at an elevation of 30 degree. the time taken by the ball to return the floor is ( g= 10 m/s2 ) 1/2 s b.1/3 s c) ¼ s d. 1s

A very broad elevator is going up vertically with a constant acceleration of 2 m/s2  . At the instant when it velocity is 4 m/s a ball is projected from the floor of the lift with a speed of 4 m/s relative to the floor at an elevation of 30 degree. the time taken by the ball to return the floor is ( g= 10 m/s2 )
 
  1. 1/2 s  b.1/3 s  c) ¼ s  d. 1s

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
The net accelaration of the particle would be (From Relative motion concept)=>
anet = g + 2 = 12 m/s^2
Hence Time of Flight = 2usin(theta)/anet  = > 2 * 4 * sin30/12 = > 4/12 = 1/3 seconds. 

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