Find h(3) and h(5)

Question

If an object is thrown straight up into the air with a velocity of 128 feet/second, then its height h(t) above the ground t seconds later is given by the formula h(t) =−16t^2+128t Find the height after 3 seconds and after 5 seconds. [Find h(3) and h(5)].

Answer 1

h(t) =-16t^2 + 128t

h(3) =-16(3)^2 + 128(3) = -144+384= 240 feet on its way up

h(5) = -16(5)^2 + 128(5) = -400 + 640 = 240 feet on its way down

maximum height is when t = 128/32 = 4 seconds

at 256 feet, when it stops going up and starts to head down