Aball 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? round to the nearest hundredth, if necessary. a) 1.75 s; 7 ft. b) 0.88 s; 43.75 ft. c) 0.88 s; 17.5 ft. d) 0.88 s; 19.25 ft

The equation would be as follows where g = gravity constant and v(o) = initial velocity, t = time in seconds and h(o) = initial height: 

h(t) = -1/2gt^2 + vt(o) +h(o) 

h(t) = -16t^2 + 28t + 0...assuming ball was somehow tossed while lying on your back. 

for example: 

h(1) = -16ft(1^2) + 28(1)ft + 0 ft = 12 ft. 

OK...revisited problem after reading your additional details: 

h = -16t^2 + 28t + 7 

h' = -32t + 28 

-32t + 28 = 0 

-32t = -28 

t = .875 seconds 

h(.875) = -16(.875^2) + 28(.875) + 7 

h(.875) = 19.25 ft 

So rounding time to nearest hundredth gives max at (.88s, 19.25ft)

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