Question

An object is 45 m above the ground when it is dropped. How fast is the object going just before it hits the ground?

Answer 1

This is a kinematics question.
$v^{2} = v_{0} ^{2} + 2g(y - y_{0}) \\ v^{2} = 0 + 2(-9.8)(0 - 45) \\ v^{2} = 882 \\ v = \sqrt{882} = 29.7 m/s$