A golfer, standing on a fairway, hits a shot to a green that is elevated 5.80 m above the point where she is standing. If the ball leaves her club with a velocity of 46.3 m/s at an angle of 36.0° above the ground, find the time that the ball is in the air before it hits the green.

Answer 1

Vo = 46.3m/s[36o].
Xo = 46.3*cos36 = 37.5 m/s.
Yo = 46.3*sin36 = 27.21 m/s.

h = (Y^2-Yo^2)/2g = (0-(27.21^2))/-19.6=
37.8 m. Above gnd.

Y = Yo + g*Tr = 0
Tr = -Yo/g = -27.21/-9.8 = 2.78 s. =
Rise time.

d = 0.5g*t^2 = 37.8-5.8 = 32 m. to fall
4.9t^2 = 32
t^2 = 6.53
Tf = 2.56 s

Tr+Tf = 2.78 + 2.56 = 5.34 s. = Time in
air.