The vertical component of feet above the groundh0 = 3 <--initial height hF = 0 <--final height 90 mi/h = 90*(88/60) ft/s = 132 ft/s v0 = 132sin(θ) --------------------------------- The horizontal component of distance along the ground. (Think of this as the shadow of the ball moving across the ground at noontime with the sun directly overhead. The shadow moves along the ground at a constant velocity.) x0 = 0 <--initial distance along ground xF = 300 <--final distance along ground 90 mi/h = 90*(88/60) ft/s = 132 ft/s <-- same as above v = 132cos(θ) ----------------------------- So we have this system of equations to solve: Solve the second equation for t and substitute in the first and get θ ≈ 16.08725911o <--answer Substitute that back and get t ≈ 2.36535260 seconds <-- you weren't asked for the time I used a TI-84 calculator to get the answer Round the answers as you were instructed. I think your dad was on the right track, but the work is so complicated, it's easy to make a mistake. Maybe I made a mistake. Edwin