At its maximum height h, the football has zero vertical velocity, so if it was kicked with initial upward speed v, then
0² - v² = -2gh
Solve this for v :
v² = 2gh
v = √(2gh)
The height y of the football t seconds after being kicked is
y = vt - 1/2 gt²
Substitute v = √(2gh), replace y = h, and solve for h when t = 3.8 s :
h = √(2gh) t - 1/2 gt²
h = √(2gh) (3.8 s) - 1/2 g (3.8 s)²
h ≈ (16.8233 √m) √h - 70.756 m
(By √m, I mean "square root meters"; on its own this quantity doesn't make much physical sense, but we need this to be consistent with √h. h is measured in meters, so √h is measured in √m, too.)
h - (16.8233 √m) √h + 70.756 m = 0
Use the quadratic formula to solve for √h :
√h = ((16.8233 √m) ± √((16.8233 √m)² - 4 (70.756 m))) / 2
Both the positive and negative square roots result in the same solution,
√h ≈ 8.411 √m
Take the square of both sides to solve for h itself:
(√h)² ≈ (8.411 √m)²
⇒ h ≈ 70.756 m ≈ 71 m
