A projectile is launched straight up from ground level with an initial velocity of 320 ft/sec when will it's height above ground be

1538 feet

Given: at time = 0, v0=+320 ft/s, [ assumed a=-g=-32.2 on earth ]

use kinematic equation for vertical projectiles,

Height,

H (t) = 1538 = v0 (t) + (1/2) at^2=320t + (1/2) (32.2) t^2

Solve for t using the quadratic formula,

with A=16.1, B=320, C=-1538:

16.1t^2+320t-1538=0

t=8.14 or t=11.74

This means that at t=8.14, the projectile reaches 1538 feet (on its way up), and at t=11.74, the projectile falls back down and reaches also 1538 feet.