An amateur rocketry club is holding a competition. there is cloud cover at 1000 ft. if a rocket is launched with a velocity of 315 ft/s, use the function 0 h (t) = ?16t + vt + h to determine how long the rocket is out of sight.

There are some errors in your given equation. It should be

h (t) = - 16t² + vt + h₀, h₀ = 0 because the rocket is launched from the ground

From physics the height of the rocket as a function of time is given by:

h (t) = vt - (1/2) gt²

Where v is the initial upward velocity and g is the acceleration due to gravity = 32 ft./sec²

Thus, h (t) = 315 t - 16 t²

Let h = 1000

1000 = 315t-16t²

16t² - 315t + 1000 = 0

Using the quadratic equation

t = (315 ± √ (3152 - (4) (16) (1000)) / 32

t = (315 ± √ (99225-64000)) / 32t = (315 ± √ (35225)) / 32

t = (315 ± 187.7) / 32

t = 127.3 / 32 or t = 502.7 / 32

thus, t = 4.0 or 15.7

In between these two times the rocket is above the clouds 15.7 - 4.0 = 11.7

Thus, the rocket is out of sight for 11.7 seconds.