A bomber is flying horizontally over level terrain, with a speed of 260 m/s rela

Question

A bomber is flying horizontally over level terrain, with a speed of 260 m/s relative to the ground, at an altitude of 3300 m. Neglect the effects of air resistance.

(a) How far will a bomb travel horizontally between its release from the plane and its impact on the ground?
Answer in m

(b) If the plane maintains its original course and speed, where will it be when the bomb hits the ground?
Which of the following? ahead of the bomb, behind the bomb, or directly above the bomb?


(c) At what angle from the vertical (less than 90°) should the telescopic bomb sight be set so that the bomb will hit the target seen in the sight at the time of release?
Answer in ° (degrees)

I can't figure this one out. Can someone please help?

Explanation / Answer

(a)
First calculate the amount of time that the bomb takes to reach the ground.

y = (1/2)g*t^2

t^2 = 2*y/g

t = sqrt(2*y/g) = sqrt(2*3300/9.8) = 25.9512887 seconds.

The x velocity remains the same, so the distance traveled in 26.73 seconds is:

x = 260* 25.9512887 = 6747.33506 m

(b)

The vertical distance is 3300m, the horizontal distance is 6747.33506 m, so the angle to the vertical is:

= ArcTan(6747.33506 /3300) degrees

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