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 th…

Question

asked Jan 12, 2020 206k views

1 Answer

Cporte · Answer 1 · 2020-01-19T13:29:23+0000

Answer:

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)

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

is the final velocity of the ball

Let's begin by finding
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)=42.01 m/s (8)

Substituting
in (3):

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

Finally:

V=26.359 m/s