Question

A projectile is fired straight upward from ground level with an initial velocity of 320 feet per second. (Assume t = 0 seconds corresponds to the time the object is fired. Use 32 feet/second2 as acceleration due to gravity.)
(a) At what instant will it be back at ground level?
(b) When will the height exceed 1584 feet? (Enter your answer using interval notation. Enter your answer in terms of seconds.)

Answer 1

(a) time up = time down
... = 320 fps / 32 ft/s^2

(b) the velocity at a given height is the same upward and downward EXCEPT for the direction

the max height is 1600 ft, which is reached in 10 sec ... the velocity is zero at this point

1584 ft is one sec below the max
... 9s < t < 11s

Answer 2

height = -16t^2 + 320t + 0

at ground level, height = 0
16t^2 - 320t = 0
16t(t - 20) = 0
t = 0 , at the start
or
t = 20

b) you want
-16t^2 + 320t > 1584
16t^2 - 320t + 1584 < 0
t^2 - 20t + 99 < 0
(t-11)(t-9) < 0

9 < t < 11 , where t is in seconds