1. **State the problem:** A box is dropped from an aeroplane 2000 m high, moving horizontally at 100 m/s. It takes 20.2 s to hit the ground. We need to find the magnitude of the displacement from the release point when the box hits the ground. 2. **Analyze vertical motion:** The box falls from a height of 2000 m under gravity. The vertical displacement is $\Delta y = 2000$ m downward. 3. **Analyze horizontal motion:** The box moves horizontally at a constant speed of 100 m/s for 20.2 s. The horizontal displacement is $$\Delta x = v_x \times t = 100 \times 20.2 = 2020 \text{ m}.$$ 4. **Calculate the magnitude of the displacement:** The displacement vector has components $\Delta x = 2020$ m and $\Delta y = 2000$ m downward. The magnitude is given by the Pythagorean theorem: $$ \text{displacement} = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{2020^2 + 2000^2}. $$ 5. **Compute the value:** $$ 2020^2 = 4,080,400, \quad 2000^2 = 4,000,000, $$ $$ \text{displacement} = \sqrt{4,080,400 + 4,000,000} = \sqrt{8,080,400} \approx 2842.3 \text{ m}. $$ **Final answer:** The magnitude of the displacement from the release point is approximately **2842.3 meters**.