Math
posted by Kiley .
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

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( (t5/2)^2  25/4) + 12
= 16(t5/2)^2 + 100 + 12
= 16(t5/2)^2 + 112
the vertex is (5/2 , 112)
so the max of h is 112 , when x = 5/2 as above
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