A ball is thrown from an initial height of 5 feet with an initial upward velocity of 40/fts. The ball's height h (in feet) after t seconds is given by the following. =h+5−40t16t2

Answer 1

you may have copied the equation a little incorrectly.

It's probably h(t) = -16t^2 + 40t + 5

then max height is found by taking the derivative and setting it =0

-32t = 40. t = 40/32 = 5/4 = 1 1/4 seconds

max height = h(1.25) = -16(5/4)^2 + 40(5/4) + 5