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