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Question

An elevator is descending with uniform acceleration . to measure the acceleration   a person in the elevator drops a coin at the moment the elevator  starts .The coin is 6ft above the floor of the elevator at the time it is dropped . The person observes that the point strikes the floor in 1s. Calculate from these data the acceleration f the elevator.


Solution

 

Let acceleration of lift be a.

acceleration of coin with respect to the lift = acceleration of coin with respesct to ground - acceleration of the lift

Or, acoin)(lift) = g - a = 9.8 m/s^2 - a = 32 ft./s^2 - a [ 9.8m = 32 ft. ]

displacement s = 6 ft.

time taken t = 1 s.

initial velocity u = 0

s = ut + 1/2 at^2 = 0*1 + 1/2 (32-a)*1*1 = 1/2 (32-a)

Therefore, 6 = 1/2 (32-a)

Hence a = 20 ft./s^2

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