An object is projected upward from ground level with an initial velocity of 160 feet per second. For what period of time is the object at least 384 feet above ground?

Answer 1

Hi Louis:

This is a standard velocity/time question. Let time to reach 384 ft be t

The relevant formula is: s = v₀t + 1/2 at²

384 = 160t - (1/2) (-32) t² [the object undergoes acceleration of -g, i.e., downward

16t² - 160t + 384 = 0

t² - 10t + 24 = 0

(t-4)(t-6) = 0

t = 4 or 6 seconds

=> it reaches height 384 in 4 seconds, and then passes the same point, on the way down, 2 seconds later

So, it remains above 384 feet for 2 seconds.