Asked by Wes
A rocket is launch from the top of an 40 foot cliff with an initial velocity of 140 feet per second, the height h of the rocket after t seconds is given by the equation h=16t^2+140t+40. How long after the rocket is launch will be 30 feet from the ground?
Answers
Answered by
Reiny
First of all the equation should say:
h = -16t^2 + 140t + 40
you want h = 30
30 = -16t^2 + 140t + 40
16t^2 - 140t - 10 = 0
t = (140 ± √20240)/32
= 8.8 or a negative, which we will reject
it would take about 8.8 seconds to reach a height of 30 ft (on its way down)
h = -16t^2 + 140t + 40
you want h = 30
30 = -16t^2 + 140t + 40
16t^2 - 140t - 10 = 0
t = (140 ± √20240)/32
= 8.8 or a negative, which we will reject
it would take about 8.8 seconds to reach a height of 30 ft (on its way down)
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