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anonymous
 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?
anonymous
 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Calculate 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Slight correction d = (1/2) * acceleration due to gravity * t ^ 2 not d = (1/2) * acceleration due to gravity * t2
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