Question

A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 14.5 m/s at an angle of 41.7° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

Answer 1

along horizontal
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initial velocity v0x = v*costheta

acceleration ax = 0

initial position = xo = 0

from equation of motion

vx = vox + ax*t

vx = vox = vo*costheta = 14.5*cos41.7 = 10.8 m/s

along vertical
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initial velocity v0y = v*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position y0 = h = 3 m

final position y = 0

from equation of motion

vy^2 - voy^2 = 2*ay*(y-y0)

vy^2 - (14.5*sin41.7)^2 = -2*9.8*(0-3)

vy = 12.3 m/s

speed before land v = sqrt(vx^2+vy^2)

v= sqrt(10.8^2+12.3^2)

v = 16.4 m/s <<<<---------------ANSWER