Question

A ball is throw horizontally from the top of a building 29.3 m high. The ball strikes the ground at a point 99.4 m from the base of the building. What is the time he ball was in motion

Answer 1

We can solve this via the kinematic equation:

$h=v_{y}t+\frac{1}{2}gt^2$

Where the vertical velocity is zero and so:

$h=0 \times t +\frac{1}{2}gt^2\\\\h=\frac{1}{2}gt^2\\\\t=\sqrt{\frac{2h}{g}}$

Since the height of the building is 29.3 meters then:

$t= \sqrt{\frac{2(29.3)}{9.8}}\\\\t \approx 2.45s$

The ball was in motion for approximately 2.45 seconds