We can solve this via the kinematic equation:
[tex]h=v_{y}t+\frac{1}{2}gt^2[/tex]
Where the vertical velocity is zero and so:
[tex]h=0 \times t +\frac{1}{2}gt^2\\\\h=\frac{1}{2}gt^2\\\\t=\sqrt{\frac{2h}{g}}[/tex]
Since the height of the building is 29.3 meters then:
[tex]t= \sqrt{\frac{2(29.3)}{9.8}}\\\\t \approx 2.45s[/tex]
The ball was in motion for approximately 2.45 seconds
