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The pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of the water t seconds after its release is modeled by $y = -16t^2 - 2t + 400$. The horizontal distance $x$ in feet between the water and its point of release is modeled by $x = 91t$. To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?

Okay, so I used the formula and got: $x = 2 \pm \frac{\sqrt{25604}}{800}$, but am not sure what to do after that.

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    $\begingroup$ Find the time to reach the ground, multiply by $91$. $\endgroup$ – André Nicolas Dec 20 '13 at 21:10
  • $\begingroup$ Im not exactly sure what you mean. $\endgroup$ – Desiree Dec 20 '13 at 21:11
  • $\begingroup$ Your quadratic formula solution is for $t$, not $x$ $\endgroup$ – Ross Millikan Dec 20 '13 at 21:15
  • $\begingroup$ So I need to find out what x equals, to plug into x=91t? $\endgroup$ – Desiree Dec 20 '13 at 21:17
  • $\begingroup$ Set $y=0$, solve the quadratic equation $-16t^2-2t+400=0$ for $t$. Discard the irrelevant negative root. Now we know how long water takes to reach the ground. Call this time $t$. We must release a horizontal distance $91t$ away, so that it reaches the ground at the fire. $\endgroup$ – André Nicolas Dec 20 '13 at 21:18
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First you have to find the time it takes for the water to hit the ground, which means solving this equation:

$$y=−16t^2−2t+400$$ Solving this (by setting $y=0$) you end up with the following for $t$ using the quadratic formula and you get the following. $t=4.937$ or $t=-5.063$.

$$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-2)\pm\sqrt{(-2)^2-4(-16)(400)}}{2(-16)}$$

Now since this is in units of time, you can disregard the negative and end up with $t=4.937$.

Then since the equation for the distance to be released is modeled by $x=91t$, just plug in for $t$ to get the distance in feet for which you should release the payload.

So you get $$t=91t=91*4.937=449.27ft$$

And that's all there is to it.

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