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A bomber is flying horizontally over level terrain at a speed of 275 $\mathrm{m} / \mathrm{s}$ relative to the ground and at an altitude of 3.00 $\mathrm{km}$ . (a) The bombardier releases one bomb. How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where is the plane relative to the bomb's point of impact when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the moment of release. At what angle from the vertical was the bombsight set?

a. 6.80$\mathrm { km }$

b. directly above the impact point

c. $66.2 ^ { \circ }$

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So the time of flight for the bomb is gonna be equaling whether we can find this out by using the kinetic equation Delta y equals v y initial times t plus 1/2 a sub y t squared. We know that this is gonna be zero the initial velocity of zero. And so we can say that the time is gonna be equaling the square root of two times delta y divided by the acceleration that one direction. And so we can say that Delta X is equaling the axe initial times t. And so we can say that then Delta X is gonna be equaling 275 meters per second and then multiplied by t, which would be the square root of two times negative three kilometers or negative, 3000 meters divided by a negative 9.8 meters per second squared. And we find that the auto acts as equaling six, 6800 meters. This would be our answer for part A. This would be the the horizontal distance traveled by the bomb. And so for part B asking us, uh, what is the plane relative to the bombs? Point of impact well because the plane, we can say that for part B, the plane is moving in the X direction at same philosophy as bombs. So when it hits the ground, it's actually keeping up with the bomb in terms of its displacement in the extraction. So it's only gonna be three kilometers from the bomb. So we can say that then the plane is 3.0 kilometers directly above the point of impact. So that would be for part B. And then to see the angle, this would simply be Fada equaling arc 10 of 6.80 kilometers divided by 3.0 kilometers. So this is equaling 66.2 degrees. This would be the required angle from of the bomb site. So this would be our final answer for part C. That is the end of the solution. Thank you for watching