Question

A ball is thrown from an initial height of 5 feet with an initial velocity of 32 ft/s. The height h (in feet) after t seconds is represented by the function h=-16t^2+32t+5. Find the maximum height of the ball.

Answer 1

Find the derivative and set equal to 0 (the speed of the object is zero when the object is at max height and s hanging direction)
-32t+32=0

So the object hits a max at t=1

Plug that into original equation and get h=21 feet