A World War II bomber flies horizontally over level terrain, with a speed of 281

Question

A World War II bomber flies horizontally over level terrain, with a speed of 281 m/s relative to the ground and at an altitude of 2.79 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. m

(b) The pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? ahead of the bomb behind the bomb directly above the bomb

(c) The bomb hits the target seen in the telescopic bombsight at the moment of the bomb's release. At what angle from the vertical was the bombsight set? ° from the vertical

Explanation / Answer

here,

the horizontal speed , ux = 281 m/s

height , h = 2.79 km = 2790 m

a)

let the time taken to hit the ground be t

h = 0 + 0.5 * g * t^2

2790= 0.5 * 9.81 * t^2

t = 23.8 s

the horizontal distance between its release and its impact on the ground , x = ux * t

x = 281 * 23.8 m

x = 6687.8 m

b)

as the horizontal speed of bomb and the plane remains the same

so the plane is above the bomb

c)

the angle from the vertical , theta = arctan(h/x)

theta = arctan(2790 /6687.8)

theta = 22.6 degree

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