Question

A dive-bomber has a velocity of 280 m/s at an angle 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.

Answer 1

To find the angle, we can use the concept of projectile motion. By breaking down the problem into horizontal and vertical components, we can calculate the angle to be approximately 52.2°.

Step-by-step explanation:

To find the angle, we can use the concept of projectile motion. Let's break the problem into horizontal and vertical components.

1. Horizontal motion: Since the horizontal velocity remains constant throughout the motion, we can use the equation distance = velocity × time. The horizontal distance traveled by the dive-bomber is 3.35 km, so we have 3.35 km = 280 m/s × t, where t is the time of flight. Solving for time, we get t = 11.96 seconds.

2. Vertical motion: We can use the equation displacement y = initial velocity × time + (1/2) × acceleration × time^2. The vertical displacement of the bomb is 2.15 km, which is equal to the initial velocity in the vertical direction (v_sin) multiplied by the time of flight (11.96 s) plus (1/2) × acceleration × time^2. Solving for v_sin, we get v_sin ≈ 397.22 m/s.

Finally, to find the angle, we can use the equation tan(angle) = v_sin / v_cos, where v_cos is the horizontal velocity of the bomb. Plugging in the values, we have tan(angle) = 397.22 m/s / 280 m/s. Taking the inverse tangent of both sides, we find that the angle is approximately 52.2°.