A punter 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.