Answer: 14 m
Explanation:
This situation is related to parabolic motion and can be solved by the following equations:
[tex]x=V_{o} cos \theta t[/tex] (1)
[tex]y=y_{o}+V_{o} sin \theta t - \frac{g}{2} t^{2}[/tex] (2)
Where:
[tex]x=30.9 m[/tex] is the horizontal distance travelled by the ball
[tex]V_{o}=18.2 m/s[/tex] is the ball's initial velocity
[tex]\theta=0\°[/tex] is the angle (it was thrown horizontally)
[tex]t[/tex] is the time
[tex]y=0 m[/tex] is the ball's final height
[tex]y_{o}[/tex] is the ball's initial height and the cliff height as well
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
Let's begin by finding [tex]t[/tex] from (1):
[tex]t=\frac{x}{V_{o}}[/tex] (3)
[tex]t=\frac{30.9 m}{18.2 m/s}[/tex] (4)
[tex]t=1.69 s[/tex] (5)
Isolating [tex]y_{o}[/tex] from (2):
[tex]y_{o}=\frac{g}{2} t^{2}[/tex] (6)
Substituting (5) in (6):
[tex]y_{o}=\frac{9.8 m/s^{2}}{2} (1.69 s)^{2}[/tex] (7)
Finally:
[tex]y_{o}=13.99 m \approx 14 m[/tex]
