A golf ball rolls off a horizontal cliff with an initial speed of 12.2 m/s. The ball falls a vertical distance of 17.3 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?

Answer 1

1. a) Dy = Viy(t) + (1/2)(g)t^2
15.5 m = 0(t) + (1/2)(9.8 m/s^2)(t^2)
t = sqrt[(15.5 m)/(4.9 m/s^2)]
t = 1.8 sec answer

b) Let V = speed of the ball just before it strikes the water.
Vix = Vfx = 11.4 m/s
Vfy = Viy + gt
Vfy = 0 + (9.8 m/s^2)(1.8 s)
Vfy = 17.64 m/s

V = sqrt[Vfx^2 + Vfy^2]
V = sqrt[(11.4 m/s)^2 + (17.64 m/s)^2]
V = 21.0 m/s

It works I just used different numbers to test.