A bomber is flying horizontally over level terrain at a speed of 300 m/s relativ

Question

A bomber is flying horizontally over level terrain at a speed of 300 m/s relative to the ground and at an altitude of 3.20 km.

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

(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?
behind the bomb
directly above the bomb
ahead of the bomb

(c) The plane has a telescopic bombsight 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?
°

Explanation / Answer

This question is about the basic principles of projectile motion. That is projectile motion is best solved by splitting into horizontal and vertical components and solving separately. The horizontal motion will remain constant. Only the vertical component will be accelerated. So, the bomb continues to travel with a horizontal component of 300 m/s. (a) Firstly calculate the time of flight (that is the time taken to hit the ground)... y = u x t + 1/2 a x t^2 (where y is the vertical displacement (3200m), u = initial velocity (0), a = acceleration due to gravity = 9.8 m/s^2). So, 3200 = 0 + 1/2 x 9.8 x t^2 t = sqrt(3200/4.9) = 25.55 s Now use this time to determine the distance travelled: x = u t = 300 x 25.55 = 7666.5 m (b) As both the plane and the bomb travel with the same horizontal component of the velocity, the bomb will always be directly underneath the plane. (c) the angle is determined from trigonometry. tan a = 7666.5 / 3200 So a = 67.3 degrees.

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