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.
