On a golf course, a green is 60m up a 10 degree slope. If the ball leaves at an angle if 20 degrees, from the ground, with what speed must the golf ball leave the club so that the ball lands on the edge of the green?
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x=v0cosθt
y=v0sinθt-1/2gt^2
conbining the two above equations [by eliminating t]
on find
y=xtanθ-gx^2/[2(v0cosθ)^2]
from which
gx^2/[2(v0cosθ)^2]=(xtanθ-y)
2(v0cosθ)^2=gx^2/(xtanθ-y)
v0=[x/(2cosθ)]*[g/(xtanθ-y)]^1/2
where
x=60*cos10=59.09 m
y=60*sin10=10.42 m
θ=20 degrees
then
v0=[59.09/(2*cos20)] * [9.8/(59.09*tan20-10.42)]^1/2
v0=29.56 m/s
y=v0sinθt-1/2gt^2
conbining the two above equations [by eliminating t]
on find
y=xtanθ-gx^2/[2(v0cosθ)^2]
from which
gx^2/[2(v0cosθ)^2]=(xtanθ-y)
2(v0cosθ)^2=gx^2/(xtanθ-y)
v0=[x/(2cosθ)]*[g/(xtanθ-y)]^1/2
where
x=60*cos10=59.09 m
y=60*sin10=10.42 m
θ=20 degrees
then
v0=[59.09/(2*cos20)] * [9.8/(59.09*tan20-10.42)]^1/2
v0=29.56 m/s
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As the ball moves upward, its vertical velocity decreases at the rate of 9.8 m/s each second. The ball’s horizontal velocity is constant. Since the green is 60 meters long and 10˚ above horizontal, let’s use the following equation to determine the final height of the green.
Final height = Length * sin θ = 60 * sin 10
This is approximately 10.42 meters. Let’s use the following equation to determine the horizontal component of the length of the green.
Horizontal component = Length * cos θ = 60 * cos 10
This is approximately 59.1 meters.
During the time that the ball is in the air, it must have a vertical displacement of approximately 10.42 meters and a horizontal displacement of approximately 59.1 meters. Since the ball leaves at an angle of 20˚ from the ground, its angle is 30˚ from horizontal. Let’s determine the vertical and horizontal components of the ball’s initial velocity.
Vertical = v * sin 30 = v * 0.5, Horizontal = v * cos 30
Let’s use the following equation to determine the time the ball is in the air.
Horizontal distance = Horizontal velocity * time
Eq#1: 60 * cos 10 = v * cos 30 * t
t = 60 * cos 10 ÷ v * cos 30
Let’s use the following equation to determine the balls vertical displacement.
Final height = Length * sin θ = 60 * sin 10
This is approximately 10.42 meters. Let’s use the following equation to determine the horizontal component of the length of the green.
Horizontal component = Length * cos θ = 60 * cos 10
This is approximately 59.1 meters.
During the time that the ball is in the air, it must have a vertical displacement of approximately 10.42 meters and a horizontal displacement of approximately 59.1 meters. Since the ball leaves at an angle of 20˚ from the ground, its angle is 30˚ from horizontal. Let’s determine the vertical and horizontal components of the ball’s initial velocity.
Vertical = v * sin 30 = v * 0.5, Horizontal = v * cos 30
Let’s use the following equation to determine the time the ball is in the air.
Horizontal distance = Horizontal velocity * time
Eq#1: 60 * cos 10 = v * cos 30 * t
t = 60 * cos 10 ÷ v * cos 30
Let’s use the following equation to determine the balls vertical displacement.
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