Question 620600: A model jet is fired up in the air from a 20-foot platform with an initial upward velocity of 68 feet per second. The height of the jet above the ground after t seconds is given by the equation h=-16t^2+68t+20, where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A model jet is fired up in the air from a 20-foot platform with an initial upward velocity of 68 feet per second. The height of the jet above the ground after t seconds is given by the equation h=-16t^2+68t+20, where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot.
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The simplest way to do this is to find the times when h = 20
One is t = 0, at the start.
h=-16t^2+68t+20 = 20
-16t^2+68t = 0
t = 0
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-16t + 68 = 0
t = 68/16 = 4.25 seconds
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t at apogee = 1/2 that since it takes the same time to ascend as to descend
t = 2.125 seconds
Sub 2.125 for t to find max height.
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