A sore golf ball tees up a frustrated golfer at the bottom of a hill with a stea

Question

A sore golf ball tees up a frustrated golfer at the bottom of a hill with a steady 20 degree incline. The golf ball then swings a golf club,striking the golfer, who flies off with an initial speed of 30 m/s at an angle of 55 degrees relative to the horizontal (so 35 degrees relative to the incline). How far up the incline has the golfer traveled when he lands?

The answer is 68.4 meters but I'm not sure how.

Our professor stated that we must first mathematically resolve the initial velocity into components using rig and that our solution must make use of one or more of the following equations:

Delta X = Vxo*t
Delta Y = Vyo*delta t -g*deltat^2/2
Vy = Vyo - g* delta t
Vy^2 - Vyo^2 = -2g*deltay

Explanation / Answer

Vx0 = 30 cos 55 = 17.2 m/s Vy0 = 30 sin 55 = 24.6 m/s h = Vy0 t - 1/2 g t^2 vertical height reached by golfer in time t (I) h = 24.6 t - 4.9 t^2 Sx = Vx0 t = 17.2 t horizontal distance golfer travels t = Sx / 17.2 h / Sx = tan 20 t = h / (17.2 * tan 20) = .16 h combining previous 2 equations h = 24.6 * .16 h - 4.9 * (.16)^h^2 = 3.94 h - .125 h^2 1 = 3.94 - .125 h h = 2.94 / .125 = 23.5 m sin 20 = h / L L = 23.5 / sin 20 = 68.7 m Hope this helps!

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