To determine the distance ahead of the target at which the plane should release the package, we can follow these steps:

1. Convert the speed to meters per second: $200 \text{ km/h} = \frac{200 \times 1000}{3600} \approx 55.56 \text{ m/s}$

2. Determine the vertical and horizontal components of the plane's velocity:

   • Vertical component $v_{y} = v \cdot \sin(\theta)$
   • Horizontal component $v_{x} = v \cdot \cos(\theta)$

   Where $\theta = 20^\circ$: $v_{y} = 55.56 \cdot \sin(20^\circ) \approx 55.56 \cdot 0.3420 \approx 19.0 \text{ m/s}$ $v_{x} = 55.56 \cdot \cos(20^\circ) \approx 55.56 \cdot 0.9397 \approx 52.2 \text{ m/s}$

3. Calculate the time of flight until the package hits the ground: We use the formula for the vertical motion under gravity: $h = \frac{1}{2} g t^2$ where $h = 1000 \text{ m}$ and $g \approx 9.81 \text{ m/s}^2$.

   Rearranging gives: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 1000}{9.81}} \approx \sqrt{203.87} \approx 14.28 \text{ s}$

4. Calculate the horizontal distance traveled during the time of flight: $d = v_{x} \cdot t = 52.2 \cdot 14.28 \approx 746.78 \text{ m}$

Thus, the plane should release the package approximately 746.78 meters ahead of the intended target.

Summary

The airplane should release the package about 747 meters ahead of its target.

Would you like further details or explanations? Here are some questions you might consider:

1. How does air resistance affect the trajectory of the package?
2. What changes if the angle of the dive is increased?
3. How would this calculation differ if the plane were at a different altitude?
4. What role does the mass of the package play in this scenario?
5. Can you explain the effect of gravity on the time of flight?

Tip: When solving projectile motion problems, always break down the motion into horizontal and vertical components for clarity.