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Grade: 12

                        

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

Answers : (3)

Raman Mishra
67 Points
							
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/s2
Since the coin falls back to its original position therefore its displacement = s=0.
Let the time taken be t.
   Using the formula  s = ut + \frac{1}{2} at^{2} ,   we get
                              0 = 20t + \frac{1}{2} (2-g)t^{2}
   Solving for t, we get  \small t= \frac{40}{g-2}
   Taking g=10m/s2 we get t= 5 seconds.
4 years ago
Prachi
29 Points
							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
nesha leda
15 Points
							
 
f=m(g+a)
ma'=m(g+a)
a'=g+a
  =10+2=12m/s^2
a'/t=v
t=a'/v
 =12/20=3/5m/s^2
 
hence the answer is  3/5 seconds
one year ago
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