The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a l20-m- high building. (a) With what speed does the elevator strike the ground? (b) For how long was it falling? (c) What was its speed when it passed the halfway point on the way down? (d) For how long was it falling when it passed the halfway point?

Given:
Initial speed of elevator,. v₀ = 0 m/s.
Acceleration due to gravity, g = 9.8 m/s²
Height from which the elevator falls, h = 120 m.
(a) From the equation of kinematics, we have

Where v represents the final speed of the object under consideration, a is its acceleration, x the distance travelled by the object, v₀ the initial speed and t the time for which the object is studied.
From the second equation, the time t is given as:

Substitute the value in first equation


Since the elevators falls freely, the acceleration (a) it has is equal to the acceleration due to gravity (g). Also, the elevator falls from height h therefore. x = h.
Substitute the given values in the equation above to have


The final speed of the elevator when it was half way down can be calculated using equation (1), but the distance x in this case would be h /2.
From equation (1), we have

For, x= h/2, v0 = 0 m a = g t/s and he above equation becomes

Rounding off to two significant figures, to have
v = 34 m/s
Therefore the final speed of the elevator when it is half way down is 34 m/s
(d) The time taken by the elevator, to fall half way down can be given using equation (2) as