A bolt comes loose from the bottom of an elevator that is moving upward at a speed of 7.52 ms and reaches the bottom of the elevator shaft in 4.90 s. How high up was the elevator when the bolt fell?

Answer 1

The original answerer misread the qustion. He figured how much further the elevator continues before the bolt lands- which has nothing to do with the question. Here's the solution to the question that was actually asked:

Acceleration (of gravity), is constant:

a = -9.81 m/s^2

Velocity of the bolt is the antiderivative of acceleration, so it is v(t) = -9.81t + C for time t and some number C. Velocity at time zero is given as 7.52, so 7.52 = 0 + C, so C = 7.52, so:

v(t) = -9.81t + 7.52

Height of the bolt is the antiderivative of velocity, so it is h(t) = -4.91t^2 + 7.52t + D for some number D. Height at time 4.90 is given as zero, so 0 = -4.91(4.90)^2 + 7.52(4.90) + D, so D = 81.04, so:

h(t) = -4.91t^2 + 7.52t + 81.04

So for t=0, height is 81.04 meters.