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avanturin [10]
2 years ago
5

100m with a constant speed 200km/hr the pilot drops abomb from the plane. determine (neplect air resistance of friction) X Q​

Physics
1 answer:
antiseptic1488 [7]2 years ago
5 0

The horizontal distance XQ traveled by the bomb is 250 m.

<h3>Distance X Q</h3>

Let the XQ be the horizontal distance traveled by the bomb.

<h3>Time for the bomb to drop from 100 m</h3>

h = vt + ¹/₂gt

Let the vertical velocity = 0

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 100 / 9.8)

t = 4.5 s

<h3>Horizontal distance traveled by the bomb</h3>

XQ = vx(t)

where;

vx is horizontal speed, = 200 km/hr = 55.56 m/s

XQ = 55.56 x 4.5

XQ = 250 m

Thus, the horizontal distance XQ traveled by the bomb is 250 m.

Learn more about horizontal distance here: brainly.com/question/24784992

#SPJ1

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Answer:

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