Question

a golfer is on the edge of a 12.5m bluff overlooking the 18th hole which is located 67.1m from the base of the bluff. she launches a horizontal shot that lands in the hole on the fly the gallery erupts in cheers. what was the ball impact velocity (velocity right before landing)

Answer 1

Answer: $26.359 m/s$

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ (2)

$V=V_{o}-gt$ (3)

Where:

$x=67.1 m$ is the horizontal distance traveled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was  a horizontal shot)

$t$ is the time

$y=0 m$ is the final height of the ball

$y_{o}=12.5 m$ is the initial height of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity

$V$ is the final velocity of the ball

Let's begin by finding $t$  from (2):

$t=\sqrt{\frac{2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$ (5)

$t=1.597 s$ (6)

Substituting (6) in (1):

$67.1 m=V_{o} cos(0\°) 1.597 s$ (7)

Finding $V_{o}$:

$V_{o}=42.01 m/s$ (8)

Substituting $V_{o}$ in (3):

$V=42.01 m/s-(9.8 m/s^{2})(1.597 s)$ (9)

Finally:

$V=26.359 m/s$