Question

# A golfer tees off from the top of a rise, giving the golf ball an initial...

A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 42.5 m/s at an angle of 30° above the horizontal.The ball strikes the fairway a horizontal distance of 185 m from the tee. Assume the fairway is level.

(a) How high is the rise above the fairway?
m

(b) What is the speed of the ball as it strikes the fairway?
m/s

Solution:

Initial velocity of the golf ball = v = 42.5 m/s

Angle of projection = 30 degrees.

Horizontal range = X = 185 m

Horizontal component of velocity = vx = 42.5 cos 30 = 36.8 m/s

Horizontal range = ( Horizontal component of velocity) (Total time ) = ( 36.8)(T) = 185

=> T = 185 / 36.8 = 5.026 s

Initial velocity along the vertical direction = vyi = 42.5 sin30 = 21.25 m/s

FInal velocity in the vertical direction = vyf =vyi -gt

= 21.25 -(9.8)(5.026)

= -28 m/s

The vertical height required is calculated using vf^2 = vi^2 +2ah

=> h = (vfy ^2 - viy^2 ) / 2g = [ (-28)^2 - (21.25)^2 ] /2*(9.8)

= 17 m

b) Speed as it strikes the fairway = vx^2 +vy^2 = ( 36.8)^2 +(-28)^2 = 46.2 m/s

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