A golfer stands 420ft(140 yd) horizontally from the hole and 50ft above the hole. Assuming the ball is hit with an initial speed of 120ft/s, at what angle should it be hit to land in the hole? Assume the path of the ball lies in a plane.

Answer 1

since the height of the ball is

y = Ho + (tanθ)x - g/(2 (v cosθ)^2) x^2
y = 50 + .1175x - .000563x^2

Just solve for θ in

50 + (tanθ)(420) - 16/(2(120 cosθ)^2) * 420^2 = 0

I get θ = 6.7°