A ball is tossed straight upward with an initial velocity of 80 ft per second from a rooftop that is 12 feet above ground level . The height of the ball in feet at time t seconds is given by h(t)= -16t^2 +80t+12 find maximum height above ground level the ball reaches

Answer 1

If you know Calculus,
h ' (t) = -32t + 80
= 0 for a max height
t = 80/32 = 5/2 or 2.5 seconds

h(2.5)= -16(2.5)^2 + 80(2.5) + 12
= 112 ft

If you don't know calculus, let's complete the square
h(t) = -16(t^2 - 5t + 25/4 -25/4 + 12
= -16( (t-5/2)^2 - 25/4) + 12
= -16(t-5/2)^2 + 100 + 12
= -16(t-5/2)^2 + 112

the vertex is (5/2 , 112)
so the max of h is 112 , when x = 5/2 as above