When will the​ projectile's height above ground be 480 ​ft? Choose the correct answer below​ and, if necessary, fill in the answer box within your choice.

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.

When will the​ projectile's height above ground be 480 ​ft? Choose the correct answer below​ and, if necessary, fill in the answer box within your choice.

A.The tile is 480 ft above ground​ twice; the first time at t=__ seconds on the way​ up, and the second time at t=__ seconds on the way down.​(Simplify your​ answers.)

B.

The projectile is 480 ft above ground once at t=__ seconds.

C.The projectile is never 480 ft above ground.

Answer 1

Using the formula s = −16t² + v₀t + s₀, we can see that v₀ = 272, s₀ = 0, and s = 480 in the context of this problem

Hence, we need to solve the equation for t:

s = −16t² + v₀t + s₀

480 = −16t² + 272t

0 = −16t² + 272t - 480

0 = (16t-240)(-t+2)

-t + 2 = 0

-t = -2

t₁ = 2

16t - 240 = 0

16t = 240

t₂ = 15

Hence, A is correct, which states, "The tile is 480 ft above ground​ twice; the first time at t=2 seconds on the way​ up, and the second time at t=15 seconds on the way down"

Hope the answers and explanations helped!