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`

Step-by-step 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 - (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{(2 y_(o))/(g)}` (4)

`t=\sqrt{(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`