Question

If a projectile is launched vertically upward from 800 ft above the ground with an initial velocity of 80 ft per second, its height in feet t seconds after launch is given by s=−16t2+80t+800
.

The projectile will take _____ seconds to return to the ground.

After _______ seconds, the projectile will be 576 feet above the ground.

Answer 1

Hello!

10 seconds to return to the ground.

7 seconds to reach 576 feet above the ground.

Find the amount of time taken to reach the ground by setting the equation equal to 0:

0 = -16t² + 80t + 800

Factor out -16 from the equation:

0 = -16(t² - 5t - 50)

Factor the terms inside of the parenthesis:

0 = -16(t - 10)(t + 5)

Find the zeros:

t - 10 = 0

t = 10

t + 5 = 0

t = -5

Time can only be positive in this instance, so the correct answer is 10 sec.

Find the time by substituting in 576 for the height:

576 = -16t² + 80t + 800

Subtract 800 from both sides:

-224 = -16t² + 80t

Rearrange:

0 = -16t² + 80t + 224

Simplify:

0 = -16(t² - 5t - 14)

0 = -16(t - 7)(t + 2)

t = 7 seconds.