Asked by callie
A golfer is attempting to hit the ball onto an 18m green over a gorge. The location
where the ball is hit from is elevated 35m above the green which is 65m away.
Assume the ball was hit at a velocity of 20m/s at 40 degrees: Questions:
1. does the ball land on the green and why?
2. What is the max speed of the ball?
3. What is the maximal height the ball reaches?
where the ball is hit from is elevated 35m above the green which is 65m away.
Assume the ball was hit at a velocity of 20m/s at 40 degrees: Questions:
1. does the ball land on the green and why?
2. What is the max speed of the ball?
3. What is the maximal height the ball reaches?
Answers
Answered by
Henry
Vo = 20m/s[40o] Above the hor.?
Xo = 20*Cos40 = 15.32 m/s.
Yo = 20*sin40 = 12.86 m/s.
1. Y^2 = Yo^2 + 2g*h = 0
h = 35 + -(Yo^2)/2g
h = 35 + -(12.86^2)/-19.6 = 43.44 m Above the green.
Y = Yo + g*Tr = 0
Tr = -Yo/g = -12.86/-9.8 = 1.31 s. = Rise time.
h = 0.5g*t^2 = 43.44
4.9t^2 = 43.44
t^2 = 8.87
t = 2.98 s. = Fall time(Tf). = Time to
fall to the green.
Dx = Xo*(Tr+Tf)=15.32m/s * (1.31+2.98)s=
65.7 m. = The required hor. distance.
Therefore, the ball lands on the green.
2. Y = Yo + g*Tf = 0 + 9.8 * 2.98 = 29.2
m/s. = Ver. component.
V = sqrt(Xo^2+Y^2)
Xo = 15.32 m/s
Y = 29.2 m/s
Solve for V.
3. h max = 43.44 m. Above the green(Prob. #1).
Xo = 20*Cos40 = 15.32 m/s.
Yo = 20*sin40 = 12.86 m/s.
1. Y^2 = Yo^2 + 2g*h = 0
h = 35 + -(Yo^2)/2g
h = 35 + -(12.86^2)/-19.6 = 43.44 m Above the green.
Y = Yo + g*Tr = 0
Tr = -Yo/g = -12.86/-9.8 = 1.31 s. = Rise time.
h = 0.5g*t^2 = 43.44
4.9t^2 = 43.44
t^2 = 8.87
t = 2.98 s. = Fall time(Tf). = Time to
fall to the green.
Dx = Xo*(Tr+Tf)=15.32m/s * (1.31+2.98)s=
65.7 m. = The required hor. distance.
Therefore, the ball lands on the green.
2. Y = Yo + g*Tf = 0 + 9.8 * 2.98 = 29.2
m/s. = Ver. component.
V = sqrt(Xo^2+Y^2)
Xo = 15.32 m/s
Y = 29.2 m/s
Solve for V.
3. h max = 43.44 m. Above the green(Prob. #1).
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