A dive bomber has a velocity of 300 m/s at an angle thete below the horizontal.

Question

A dive bomber has a velocity of 300 m/s at an angle thete below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from release of the bomb to the target is 3.15 km. Find the angle theta.

Explanation / Answer

Horizontal initial velocity = 300cos(theta) Vertical initial velocity = 300sin(theta) Horizontal acceleration = 0 Vertical acceleration = 9.81m/sec^2 In vertical Direction: v^2 = u^2 + 2as =>v^2 = (300sin(theta))^2 + 2*9.81*2150 v = u + at =>sqrt((300sin(theta))^2 + 2*9.81*2150) = 300sin(theta)+9.81t =>(sqrt((300sin(theta))^2 - 2*9.81*2150) - 300sin(theta))/9.81 = t In Horizontal Direction: x = 300cos(theta)*t = 300cos(theta)(sqrt((300sin(theta))^2 - 2*9.81*2150) - 300sin(theta))/9.81 =>300cos(theta)(sqrt((300sin(theta))^2 - 2*9.81*2150) - 300sin(theta))/9.81 = sqrt(3150^2 - 2150^2) = 2302.172887 Solve for theta.

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