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  • 5 years ago

A bomber flies horizontally with a speed of 284 m/s relative to the ground. The altitude of the bomber is 1470 m and the terrain is level. Neglect the effects of air resistance. The acceleration of gravity is 9.8 m/s^2. How far from the point vertically under the point of release does a bomb hit the ground?

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  1. anonymous
    • 5 years ago
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    Calculate the total time it will take the bomb to hit the ground using one of the Kinematic Equations for Constant Acceleration such as d = V(initial) * time + (1/2) * acceleration * t^2. Since the bomb is being dropped from a bomber which is not changing altitude V(initial) of the bomb is zero. d = (1/2) * acceleration due to gravity * t2. t = square root of 2*a*d or square root of 2 * 9.8 m/s^2 * 1470m. Using the answer you find for t you can calculate the distance the bomb travels using the velocity of the bomber at time of release. Distance is velocity * time so your final solution should be 284 m/s * square root of (2 * 9.8 * 1470).

  2. anonymous
    • 5 years ago
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    Slight correction d = (1/2) * acceleration due to gravity * t ^ 2 not d = (1/2) * acceleration due to gravity * t2

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