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a body is throw upin a lift with a velocity 'u' relativeto the lift. it returns to the floor of lift in time t. then the upward acceleration to the lift is
Hi Sree,
Let the upward accelaration of the lift is a' .
Now for an observer in the lift , a pseudo force ma' will be acting on the body ( of mass m ) in the direction opposite to the direction of motion of accelaration of lift.
Thus forces acting on body will be : ma' downwards and mg downwards.
Thus Net force = MA = ma' + mg where A is net accelaration of the body. A is downwards.
Now, using equation of motion,
0 = ut + 1/2( - A )t2
1/2At2 = ut
At = 2u
A = 2u/t
Now a' + g = A
a' + g = 2u/t
Thus a' = 2u/t - g is upward accelaration of the lift.
Please feel free to ask as many questions as you want.
Puneet
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