Answer: [tex]26.359 m/s[/tex]
Explanation:
This problem 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{1}{2}gt^{2}[/tex] (2)
[tex]V=V_{o}-gt[/tex] (3)
Where:
[tex]x=67.1 m[/tex] is the horizontal distance traveled by the golf ball
[tex]V_{o}[/tex] is the golf ball's initial velocity
[tex]\theta=0\°[/tex] is the angle (it was a horizontal shot)
[tex]t[/tex] is the time
[tex]y=0 m[/tex] is the final height of the ball
[tex]y_{o}=12.5 m[/tex] is the initial height of the ball
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]V[/tex] is the final velocity of the ball
Let's begin by finding [tex]t[/tex] from (2):
[tex]t=\sqrt{\frac{2 y_{o}}{g}}[/tex] (4)
[tex]t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}[/tex] (5)
[tex]t=1.597 s[/tex] (6)
Substituting (6) in (1):
[tex]67.1 m=V_{o} cos(0\°) 1.597 s[/tex] (7)
Finding [tex]V_{o}[/tex]:
[tex]V_{o}=42.01 m/s[/tex] (8)
Substituting [tex]V_{o}[/tex] in (3):
[tex]V=42.01 m/s-(9.8 m/s^{2})(1.597 s)[/tex] (9)
Finally:
[tex]V=26.359 m/s[/tex]
