An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 300 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. (Round the final answer to the nearest whole number.)

The horizontal velocity of the water is constant and equal to the speed of the plane, since there is not any horizontal force (by neglecting the air drag).

[tex]V_x=180mi/h=\dfrac{d}{t}[/tex]

Solving for d:

[tex]d=180mi/h\cdot t[/tex]

The initial vertical velocity is zero, because the plane is flying horizontally.

The time, t, is determined by the altitude (h), using the equation with initial vertical velocity zero.

[tex]h=g\cdot t^2/2[/tex]

Solve for t, substitute and compute:

[tex]t=\sqrt{\dfrac{2h}{g}}\\ \\\\t=\sqrt{\dfrac{2(300ft)}{32.174ft/s^2}}=4.3184s[/tex]

Before calculate d, convert the speed to ft/s.

[tex]180mi/h\times 5280ft/mi\times 1h/3,600s=264ft/s[/tex]

Now calculate d, using the first equation:

[tex]d=180mi/h\cdot t\\\\d=264ft/s\times4.3184s=1140ft[/tex]

The answer is rounded to the nearest whole number by specifications of the problem.

batolisis batolisis · Answer 2 · 2022-02-24T12:45:46+01:00

The distance ( d ) at which the pilot should release the water is : 1140 ft

Given data :

Velocity ( Vo ) = 180 miles/h = 164 ft/s

Altitude = 300 ft

Determine the distance ( d )

let's place the origin at point A

For a vertical motion ( y ) = - 1/2 gt²

At point B

- 300 ft = - 1/2 ( 32.2 ) t² ( g = 32.2 ft/s² )

therefore ; t = 4.31666 secs

while

For a uniform horizontal motion ( x ) = 0 + ( Vx ) ot

Therefore the distance d at which the pilot should release the water to hit the fire at B

I.e. B : d = 264 ft/s / 4.31666 secs

= 1140 ft

Hence we can conclude that The distance ( d ) at which the pilot should release the water is : 1140 ft

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