Question

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launches a horizontal shot that lands in the hole on the fly. The gallery erupts in cheers.

Answer 1

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 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 = m is the horizontal distance travelled 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 is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

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

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

$t=1.597$s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$-------------------(4)

Step 2:  Finding $V_{o}$:

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)  

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

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566  

V=26.359 m/s