Posted by **Kiley** on Sunday, February 10, 2013 at 9:16pm.

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

- Math -
**Reiny**, Sunday, February 10, 2013 at 9:51pm
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

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