Question

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone's velocity just before it strikes the ground?

Answer 1

This is a free-fall problem. This can be answered using one of the free-fall equations:

V^2 = 2gh or V = √(2gh)

Where V = velocity ; h = total height (given as 318m) ; g = acceleration due to gravity (as this is on Earth, let us use 9.8 m/s^2)

With the given values, we can substitute it into the equation directly like so:

V = √(2gh)
V = √(2 x 9.8 x 318)
V = √(6232.8)
V = 78.94808 or approximately 78.95 m/s

Therefore the stone's velocity just before hitting the ground is 78.95 m/s.