Apunter on westview high school's football team kicks a ball straight up from 5 feet above the ground with an initial upward velocity of 45 feet per second. the height of the ball above ground after t seconds is given by the equation h = –16t^2 + 45t + 5, where h is the height of the ball in feet and t is the time in seconds since the punt. what is the maximum height of the ball, to the nearest foot?

I hope this is right. 

h(x) = -16t^2 + 45t + 5

Integrate function (make y the subject)
h'(x) = -32t + 45
y = -32t + 45
-45 = -32t
-45/-32 = -32t/-32
1.40625 = t
t = 1.40625 

Plug t value back into the original function
h(1.40625) = -16*1.40625^2 + 45*1.40625+ 5 = 36.64065

The maximum height of ball above ground after t seconds is thus 37 feet. (to nearest feet)

Note that you check this by graphing the function on your calculator and looking at the y coordinate at the maximum of the function.