Question

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?

Answer 1

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).