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a person in an elevator accelerating upwards with an acceleration of 2m/s2 tosses a coin vertically upwards with a speed of 20 m/s after how much time will the coin fall back into his hand?


4 years ago

							Taking the elevator as the reference frame, we have:   Initial velocity of the coin wrt the elevator = u = 20m/s   Acceleration of the coin wrt the elevator = a = (2-g)m/s2Since the coin falls back to its original position therefore its displacement = s=0.Let the time taken be t.   Using the formula   ,   we get                                 Solving for t, we get     Taking g=10m/s2 we get t= 5 seconds.

4 years ago
							Taking the elevator as a reference frame. Now since elevator is a non-inertial frame pseudo force is applied. Since elevator is accelerating upwards therefore pseudo force acts in downward direction By drawing the F.B.D of thw coin, summation of force on y-axis give F = ma(psuedoforce) + mg(weight of coin)mA = m(g+a)A = g+aA = 10+2A = 12m/s^2 . Hence new acceleration of coin is 12m/s^2.Now time taken by the coin to come back in the hand is 2t and t =v/ATherefore 2 x 20/12= 40/12t = 3.33 s

3 years ago
							 f=m(g+a)ma'=m(g+a)a'=g+a  =10+2=12m/s^2a'/t=vt=a'/v =12/20=3/5m/s^2 hence the answer is  3/5 seconds

2 years ago
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