Posted by Jin on Monday, March 26, 2012 at 6:44pm.
A ball is thrown into the air with an upward velocity of 28 ft/s. Its height (h) in feet after t seconds is given by the function h = –16t² + 28t + 7. How long does it take the ball to reach its maximum height? What is the ball’s maximum height?
How do I solve this?

algebra  Anonymous, Monday, March 26, 2012 at 7:14pm
the vertex of a parabola is at
x = b/2a = 28/32 = 7/8
h(7/8) = 19.25

algebra  Reiny, Monday, March 26, 2012 at 7:21pm
If you know Calculus, this is easy
dh/dt = 32t + 28 = 0 for a max of h
32t = 28
t = 28/32 = 7/8 = .875
h = 16(49/64) + 28(7/8) + 7 = 77/4 = 19.25 ft
if you don't know Calculus, you must complete the square
h = 16(t^2  28/16t + ...) + 7
= 16(t^2  (7/4)t + 49/64  49/64) + 7
= 16( (t7/8)^2  49/64) + 7
= 16(t7/8)^2 + 49/4 + 7
= 16(t7/8)^2 + 77/4
the vertex of this quadratic is (7,8 , 77,4)
so there is a max of h of 77/4 or 19.25 when h = 7/8
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