Question

A golf ball rolls off a horizontal cliff with an initial speed of 12.7 m/s. The ball falls a vertical distance of 16.0 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

along vertical

initial velocity voy = 0

initial position yo = 16.0 m

fina position y = 0

acceleration ay = -g = -9.81 m/s^2

from equation of motion

y - yo = voy*t + (1/23)*ay*t^2

0 - 16 = 0 - (1/2)*9.81*t^2

time t = 1.80 s ,<<<-------answer

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(b)

along horizontal

initial velocity vox = v = 12.7 m/s

acceleration ax = 0

vx = vox + ax*t = 12.7 m/s

alongvertcial

vy = voy + ay*t

vy = 0 - 9.81*1.8

vy = -17.7 m/s

therefore speed v = sqrt(vx^2 + vy^2)

v = sqrt(12.7^2 + 17.7^2 = 21.8 m/s = 22.0 ( 2 sig figures) <<<-----answer