Methods of Calculus give the soccer ball's position in coordinates (x,y) on a two-dimensional graph (at t seconds after kicking) as: x = (v0cos θ)t; y = (v0sin θ)t - 0.5gt2 where v0 is the velocity of the ball at the instant it is kicked and θ is the instantaneous angle of the ball's straight-line path with the unsloped ground immediately after kicking.
The acceleration (or instantaneous rate of change of velocity with respect to time) due to gravity is "g" equal to 32 feet per second, every second, or g = 32 feet/second2.
From x = (v0cos θ)t, write t = x/(v0cos θ) and put this expression for t into y = (v0sin θ)t - 0.5gt2.
y = (v0sin θ)(x/v0cos θ) - 0.5(32)(x/v0cos θ)2 will simplify to y = xtan θ - 16x2/(v02cos2θ).
The vertical or "straight-up" velocity of the ball is given by Calculus as v0sin θ - 32t; when the ball is at the top of its path (8 feet) and is just starting to descend, v0sin θ - 32t = 0 because the ball is no longer climbing and has no velocity in the y-direction. Then at maximum height (8 feet), v0sin θ = 32t or
t = (v0sin θ)/32.
Now replace t with (v0sin θ)/32 in the equation y = (v0sin θ)t - 0.5gt2; also replace y with 8 and g with 32.
8 = (v0sin θ)(v0sin θ/32) - 0.5(32)(v0sin θ/32)2 or 8 = (v02sin2θ)(1/32 - 16/322). This gives v02sin2θ = 512 and v0sin θ = √512.
To find the time of the ball's landing, set y = t(v0sin θ - 0.5gt) = 0; t = 2v0sin θ/g at landing or t = (2√512)/32 equal to √2 seconds at landing.
From x = (v0cosθ)t, obtain 30 = (v0cos θ)√2 or 30/√2 = v0cos θ.
Now tan θ can be found by (v0sinθ)/(v0cos θ) equal to √512/(30/√2) which gives 16/15. This gives θ = arctan (16/15) ≈ 46.85°.
Finally, x = (v0cosθ)t or 30 = v0cos (arctan 16/15)√2 which gives v0 as 31.01612484 feet/second.
The quadratic equation sought is then y = xtan θ - 16x2/(v02cos2θ) or (16/450)x2 - (16/15)x + y = 0; y as a function of x is y = (16/15)x - (8/225)x2.
