Question

A dive bomber has a velocity of 240 m/s at an angle θ 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 the point of release of the bomb to the target is 3.35 km. Find the angle θ in degrees.

Answer 1

To find the angle θ at which the bomb is released, the Pythagorean theorem and trigonometric functions are used with the given displacement and velocity components. The angle of release is approximately 32.62 degrees below the horizontal.

Step-by-step explanation:

To find the angle θ at which the dive bomber releases the bomb, we can make use of trigonometric relationships in projectile motion. The known values are the velocity of the aircraft (240 m/s), the altitude from which the bomb is released (2.15 km or 2150 m), and the magnitude of the displacement from the release point to the target (3.35 km or 3350 m).

Using the Pythagorean theorem for the horizontal and vertical components of the displacement, we can write:
tan(θ) = vertical/horizontal
tan(θ) = 2150 m / (3350 m)
θ = tan-1(2150 m / 3350 m)
θ ≈ tan-1(0.6418)
θ ≈ 32.62°

Therefore, the angle θ below the horizontal at which the dive bomber releases the bomb is approximately 32.62 degrees.