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 = √(882) = 29.7 m/s`