(C) When will the​ projectile's height above ground be less than or equal to 480 ft?

Question

If an object is given an initial velocity straight upward of v₀ feet per second from a height of s₀ ​feet, then its altitude s after t seconds is given by the formula s=−16t2+v₀t+s₀. A projectile is launched straight up from ground level with an initial velocity of 272 ​ft/sec.

(C) When will the​ projectile's height above ground be less than or equal to 480 ft?

Answer 1

480 >= −16t2+272t+0

30 >= -t^2 + 17t

-t^2 + 17t - 30 <= 0

t^2 - 17t + 30 >= 0

(t - 15)((t-2) >= 0

15 and 2 are the points to check: - inf to 2, 2 to 15, 15 to infinity

try 0: 30 >= 0. Yes. t <= 2

2 to 15, try 10. 10^2 - 17*10 + 30 >= 0?

100 - 170 + 30 = -40, not >= 0. NO. 2 <= t <= 15 is out.

15 to infinity, try 100: 100^2 - 17*100 + 30 >= 0

10000 - 1700 + 30 = 8330 >= 0 YES. t >= 15

The projectile is 480 feet or less when t <= 2 and when t >= 15 seconds.