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 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.
Given:
Distance between the coin and the floor of the elevator before the coin is dropped = 6 ft
Let, a= the acceleration of the elevator.
It is given that the coin reaches the floor in 1 second. This means that the coin travels 6 ft distance.
The initial velocity is u for the coin and 0 for the elevator.
Using the equation of motion, we get:
Equation for the coin:
=
Here,
a‘= g − a (a‘ is the acceleration felt by the coin.)
g = Acceleration due to gravity
g = 9.8 m/s2 = 32 ft/s2
on substituting the values, we get:
=
=
Therefore, we can write:
Hence, the acceleration of the elevator is 20 ft/s2.