When it comes to horizontal projectiles, the formula for time is:
[tex]t = \sqrt{ \frac{2dy}{g} } [/tex]
Where:
dy = vertical distance or height
g = acceleration due to gravity
t = time
Based on the problem, we know that the height at which the tennis ball was thrown is 78.4 m and the acceleration due to gravity is a constant 9.8m/s2. All you need to do is input that into our equation:
[tex]t = \sqrt{ \frac{2dy}{g} } [/tex]
[tex]t = \sqrt{ \frac{(2)(78.4m)}{9.8m/s^{2}} } [/tex]
[tex]t = \sqrt{ \frac{156.8m}{9.8m/s^{2}} } [/tex]
[tex]t = \sqrt{16s^{2}} [/tex]
[tex]t = 4s[/tex]
The time it would take is 4 seconds.
