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An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 450 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. The distance at which the pilot should release the water so that it will hit the fire at B is ft.

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raphealnwobi

Answer:

1399.2 ft

Explanation:

The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s

[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]

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