Question 821684: A rocket is fired upward with an initial velocity of 168 ft/s. Using the formula h = rt-16t^2 determine when the rocket is 360 ft high and when does it crash.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
h(t) = -16t^2 + 168t
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at 360 feet:
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h(t) = 360 = -16t^2 + 168t
-16t^2 + 168t - 360 = 0
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the above quadratic equation is in standard form, with a=-16, b=168, and c=360
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-16 168 -360
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the two real roots (i.e. the two x-intercepts), of the quadratic are:
t= 3 sec
t= 7.5 sec
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answer 1:
the rocket is at 360 ft of altitude at 3 seconds after launch, and then again at 7.5 seconds after launch
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at 0 feet:
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h(t) = 0 = -16t^2 + 168t
-16t^2 + 168t = 0
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the above quadratic equation is in standard form, with a=-16, b=168, and c=0
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-16 168 0
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the two real roots (i.e. the two x-intercepts), of the quadratic are:
t= 0 sec
t= 10.5 sec
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answer 2:
the rocket crashes to earth at 10.5 seconds after launch
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Solve quadratic equations, quadratic formula:
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Solve systems of linear equations up to 6-equations 6-variables:
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