posted by walter hughes .
the trajectory, of the projectile satisfies the following equation,wher g is the acceleration due to gravity, y=mx-g/2v^2(1+m^2)x^2 where y as the vertical rise, and x asthe horizontal component of the displacement. A football is kicked with initial velocity v= 80 fett per second on a initial line of y=1/2x,where g=32ft/s^2. By considering horizontal tangents to the trajectory,find how high the football will rise.
so, plug in your numbers. You don't say what m is, but I assume it's the slope of the trajectory at kickoff
y = (1/2)x - 32/(2*80^2)(1+1/4)x^2
= 1/320 (160x-x^2)
this is a parabola with vertex at (80,20)
so, knowing that, what's the max height?