Answer:
The water should be released a distance of 1289.88 feet before the plane is on top of the bush fire.
Explanation:
Let's first find the time required for the water to fall down 390 ft onto the bush fire.
Initial Vertical Speed of water = 0
Distance to be covered = 390 ft
Acceleration due to gravity = 32.2 ft/s^2
[tex]s=u*t+\frac{1}{2} (a*t^2)[/tex]
[tex]390=0*t+0.5(32.2*t^2)[/tex]
t = 4.92 seconds
Thus the water should be dropped 4.92 seconds before the plane is over the bush fire. Now we can also find the distance d at which the pilot should release the water:
d = Speed of plane * time
Speed of plane = 180 / (60 * 60) = 0.05 mile/second
d = 0.05 * 4.92 = 0.246 miles OR 1289.88 feet
Thus, the water should be released a distance of 1289.88 feet before the plane is on top of the bush fire.
