A golfer is on the edge of a 15 m bluff overlooking the 18th hole

which is located 70 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.

What was the ball's impact velocity (velocity right before

Answer:reg

V = 40m/s

Explanation:

The ball was launched horizontally, its initial vertical velocity is 0 m/s.

We use the following equation to determine the time to fall 15m

S = ut + ½ at², u = 0

15 = ½ * 9.8 * t²

15 = 4.9t²

t = √15/4.9

t = 1.7496secs

This is approximately 1.75secs. During this time, the ball moves a horizontal of 70m. We'll use the equation to determine the ball's horizontal velocity.

S = V*t

70 = V*1.75

V = 70/1.75

= 40m/s.