Stone Hit Moving Drone Simulation

Knowledge Points:

Projectile Motion: Stone moves along a parabolic trajectory. Horizontal speed is constant; vertical speed is affected by gravity.
Maximum Height & Landing Point: Maximum height occurs when vertical velocity is zero. Landing point occurs when stone returns to ground or drone height.
Relative Motion: Drone moves horizontally at a constant speed. Collision happens when the stone's and drone's positions coincide.

Key Formulas:

Stone's horizontal position: $x(t) = v_0 \cos(\theta) \cdot t$
Stone's vertical position: $y(t) = v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2$
Maximum height: $y_{max} = \frac{(v_0 \sin(\theta))^2}{2g}$

Practice Question:

A stone is launched from the bottom left corner with an initial velocity of v₀ = 12 m/s at an angle of 45°. The drone moves horizontally left from position x = 12, y = 5 at a speed of 0.2 units per 0.05s. Will the stone hit the drone?

Answer & Explanation:

Answer: B. No

Solution Breakdown:

**1. Calculate the Maximum Height**
The initial vertical velocity component of the stone is:
$v_{y0} = v_0 \sin(\theta) = 12 \cdot \sin(45°) \approx 8.485 \text{ m/s}$
The time required to reach the highest point is:
$t_{max\_height} = \frac{v_{y0}}{g} = \frac{8.485}{9.8} \approx 0.866 \text{ s}$
Substitute this time into the vertical position formula to calculate the maximum height:
$y_{max} = v_{y0} \cdot t - \frac{1}{2} g t^2 = 8.485 \cdot (0.866) - 0.5 \cdot (9.8) \cdot (0.866)^2 \approx 3.68 \text{ m}$
**2. Determine if it will hit**
The drone's height is **5 meters**.
The stone's maximum height is **3.68 meters**.
Because the stone's maximum height is lower than the drone's height (3.68 m < 5 m), the stone will never reach the drones altitude and therefore cannot hit it.