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3 years ago

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A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

f 3.24 km. The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance.

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

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