Analyzing an Aircraft's Bomb Release Trajectory

Problem Overview

An aircraft dives at a speed of 290 kilometers per hour at a 30-degree angle below the horizontal. A bomb is released, and the horizontal distance between the release point and where the bomb strikes the ground is 700 meters. The tasks are to determine how long the bomb remains in the air and the height from which it was released.

Key Concepts

1. Projectile Motion: Separating the motion into horizontal and vertical components.
2. Time of Flight: Calculating the time the bomb remains in the air.
3. Vertical Displacement: Determining the height from which the bomb was released.

Step-by-Step Solution

Part A: Time of Flight

• • Horizontal Velocity Calculation:Convert 290 km/h to meters per second: 290 km/h = 80.56 m/s.
  • Calculate horizontal component of velocity (Vx): V cos theta = 80.56 cos(-30) = 69.76 m/s.
  • Use the formula Time = Distance / Velocity.
  • Time (t) = 700 meters / 69.76 m/s ≈ 10.03 seconds.
  • Result: The bomb is in the air for approximately 10.03 seconds.

Part B: Height of Release

• • Vertical Velocity Calculation:Calculate vertical component of velocity (Vy): V sin theta = 80.56 sin(-30) = -40.28 m/s (negative as the angle is below the x-axis).
  • Vertical Displacement Equation:Use the equation y - y0 = Vy * t - 1/2 * g * t².
  • Rearrange to find y0 (initial height): y0 = y - Vy * t + 1/2 * g * t².
  • Assuming final y (ground level) = 0.
  • Substitute values: y0 = 0 - (-40.28 m/s * 10 s) - 1/2 * 9.81 m/s² * 10².
  • Result: Initial height y0 ≈ 897 meters above the ground.

Conclusion

• The bomb released from the aircraft remains in the air for approximately 10.03 seconds.
• The height from which the bomb was released is approximately 897 meters above the ground. This solution illustrates the principles of projectile motion in a real-world context, separating the motion into horizontal and vertical components to solve complex problems.