a cannonball's initial upward velocity is 128 feet per second. at what time will it be 192 feet above the ground

Answer 1

If the cannon was fired from the ground, the equation for height vs. time, with height in feet and time in seconds, is

h(t) = -16t² + 128t

Set h(t) equal to 192 and solve for t.

-16t² + 128t = 192

-16t² + 128t - 192 = 0

Dive through by -16 and get

t² - 8t + 12 = 0

(t - 6)(t - 2) = 0

Note you get two solutions: t=6 sec and t=2 sec

It reaches 192 feet on the way up in 2 seconds, and reaches 192 ft on the way down in 6 seconds.