y=-30x^2 + 5x
max height is when y'=0 =-60x +5, x=5/60 = 1/12th of a second
max y = -30(1/12)^2 +5/12 =-30/144 + 5/12
=5/12 -15/72
=5/12 -5/24
= 10/24 -5/24
= 5/24 feet high
= 2 1/2 inches
it hits the ground when y=0=-30x^2+5x
-5x(6x-1), x= 1/6 of a second
if it reaches 25 meters horizontally, that happens at any time less than or= 1/6 of a second
less than or = 150 meters per second is the horizontal velocity, which seems impossible.
so it never reaches 25 meters vertically or horizontally
it reached max height in half that time, x= 1/12 of a second
there is no constant because the constant = the initial height of the ball and apparently it is on the ground when kicked
this problem might be more realistic if the equation were y=-3.0x^2 +5x, moving the decimal point one digit left