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.

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