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.

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2016-09-25 20:59:55

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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