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? 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)