Drone Package Drop with Horizontal Wind

We model a drone flying horizontally and dropping a package that then falls under gravity while being pushed back by wind.

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Problem Setup

• Drone flies horizontally at a constant speed:

$$v_{d} = 20\ \text{m/s}$$

• Drone height (release height):

$$h = 80\ \text{m}$$

• After release, the package is affected by:
  • Gravity: $g \approx 9.8\,\text{m/s}^2$
  • A horizontal wind in the opposite direction at speed:

$$v_{w} = 5\ \text{m/s}$$

• The wind only acts after the package is released.
• Take horizontal motion to the right as positive. The drone moves right, wind pushes the package left.

We assume:

• No air resistance besides the constant effective wind speed.
• The person stands somewhere on the ground, waiting to receive the package.

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Equations of Motion

1. Time of Flight

The vertical motion of the package after release is a simple free fall:

• Initial vertical velocity: $v_{y0} = 0$
• Initial vertical position: $y_0 = h = 80$ m (measured above ground)

Using

$$y(t) = y_0 - \frac{1}{2} g t^2,$$

we set $y(T) = 0$ at impact (ground), so:

$$0 = h - \frac{1}{2} g T^2 \Rightarrow T = \sqrt{\frac{2h}{g}}.$$

With $h = 80 \text{ m}$, $g = 9.8 \text{ m/s}^2$:

$$T = \sqrt{\frac{2\cdot80}{9.8}} \approx \sqrt{16.33} \approx 4.04\ \text{s}.$$

So the package is in the air for about 4.0 seconds.

2. Horizontal Motion (Drone)

While flying, the drone moves at constant horizontal speed $v_d$:

$$x_{drone}(t) = x_{0} + v_{d} t.$$

We often choose $x_0 = 0$ at the release point, so after release the drone is just moving away, but for targeting we care about where the package lands relative to that release point.

3. Horizontal Motion (Package with Wind)

At release, the package inherits the drone's horizontal speed. Let the moment of release be $t = 0$ and horizontal position $x=0$. Then the initial horizontal velocity of the package is:

$$v_{x0} = v_{d} = 20\ \text{m/s}.$$

Now the wind acts in the opposite direction with speed 5 m/s. A simple way to model this (in this visualization) is to treat it as giving the package an effective constant horizontal velocity:

$$v_{x,\text{eff}} = v_{d} - v_{w} = 20 - 5 = 15\ \text{m/s}.$$

So:

$$x_{\text{pkg}}(t) = v_{x,\text{eff}} t = 15 t.$$

At the time it hits the ground $t = T$:

$$x_{\text{impact}} = v_{x,\text{eff}} T = (v_d - v_w)\sqrt{\frac{2h}{g}}.$$

This is the horizontal distance from the release point to where the package lands.

To deliver the package to a person standing at some distance $D$ from the point directly under the drone at release, the drone must release the package when that vertical line is located so that:

$$D = x_{\text{impact}} = (v_d - v_w)\sqrt{\frac{2h}{g}}.$$

In our visualization, you can change:

• Drone speed $v_d$
• Wind speed $v_w$
• Height $h$
• Gravity $g$
• Person distance from release point $D$

and see how the trajectory changes and whether the package lands on the target.

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What the Visualization Shows

On the dark background, we display:

1. Ground line and vertical scale.
2. The drone moving horizontally at a chosen height.
3. The package trajectory as a bright parabolic path:
   • Horizontal motion: constant speed $v_d - v_w$
   • Vertical motion: accelerated downward with gravity.
4. A person standing on the ground at a chosen horizontal distance from the point under the drone at the release moment.
5. A visual indication (color change / glow) when the landing point is close enough to the person (successful delivery).

The simulation loop uses a parameter that lets you:

• Watch the drone fly along.
• Trigger or view the effective release moment.
• Observe how the projectile curve reaches the ground.

Mathematically, the key relationship is the dependence of horizontal range on the effective horizontal speed and fall time:

$$\text{Range} = (v_d - v_w)\sqrt{\frac{2h}{g}}.$$

By experimenting with the sliders, you see how higher height $h$ or higher drone speed $v_d$ increases range, while stronger opposing wind $v_w$ reduces it.