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( (t-7/8)^2 - 49/64) + 7

= -16(t-7/8)^2 + 49/4 + 7

= -16(t-7/8)^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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