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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 coin striked the floor in 1 second.Calculate 4m these data the acceleration of the elevator.
So the coin is dropped at the instant when elevator starts , u = 0Time taken by coin to rech floor is 1 sec so the distance covered by coin will be S=ut+0.5*gt^2 eqn 1 S=0*1+0.5*9.8*1=4.9 m Now as given in problem the distance of coin before dropping and floor was 1.82m(6ft) which implies that the elevator moved 4.9-1.82=3.07m during the time coin strikes floorsubstituting S' in equation 1 we get S=ut+0.5*at^2 3.07=0*1+0.5*a*1 a=6.14m/sec^2 (acceleration of the elevator)
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