Question

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 = -16t2 - 2t + 500. The horizontal distance x in feet between the water and its point of release is modeled by x = 91t ft. At what horizontal distance from the fire should the pilot start releasing the water in order to hit the target?

Answer 1

We'll first need to find the time required for the water to fall all the way to the ground. We can do this by setting the first equation, y, equal to 0:
\[0=-16t^2 -2t +500\] If you graph this, you'll find the solution is 5.44 seconds. Now we plug this time in to the second equation to find the horizontal distance the water will travel, which is our answer:
\[x=91t=91(5.44)=495 ft\]

Answer 2

Step 1 Use the first equation to determine how long it will take the water
to hit the ground. Set the height of the water equal to 0 feet, and use the
quadratic formula to solve for t.
\[y=-16t^2 -2t+500\]
Put y=0
therefore
\[0=-16t^2 -2t+500 \]
\[-16t^2 -2t+500 = 0 \rightarrow 16t^2 +2t-500 = 0\]
Comparing \[16t^2 +2t-500 = 0\] by \[at^2 +bt+c= 0\] (i.e. general form of quadratic eq) we find)
a= 16, b= 2, c=-500
Now by quadratic eq formula we have \[t= \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
Now substituting the values of a, b, c in the above formula.
\[t= \frac{ -2 \pm \sqrt{2^2-4 \times 16 \times (-500)} }{ 2 \times 2}\]
\[t= \frac{ -2 \pm \sqrt{4+32000} }{32}=\[\frac{ -2 \pm \sqrt{32004} }{32}
=\frac{ -2 \pm 178.89 }{32}=\frac{ -2 \pm 179 }{32}\] \[[since \sqrt{32004}=178.89 \approx 179]\]
\[t=\frac{ -2 - 179 }{32} or t=\frac{-181 }{32} =-5.65\]
or
\[ t=\frac{ -2 +179 }{32} = \frac{177 }{32} =5.53\]
Since time can not be negative, so the water lands on the target about
5.5 seconds after it is released.

Step 2 Find the horizontal distance that the water will have traveled in
this time.
x = 91t
x = 91 (5.5) Substitute 5.5 for t.
x = 500.5 Simplify.
The water will have traveled a horizontal distance of about 500 feet.
Therefore, the pilot should start releasing the water when the horizontal
distance between the helicopter and the fire is 500 feet.
Check Use substitution to check that the water hits the ground after about
5.53 seconds.
\[y=-16t^2 -2t+500 =y=-16(5.53)^2 -2(5.53)+500\]
i.e. \[y \approx -0.3544 \approx 0\]
The height is approximately equal to 0 when
t = 5.53.