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 angle between the horizontal and the displacement vector of the box from the release point. 2. **Identify known values:** - Initial horizontal velocity, $v_x = 100$ m/s - Time of flight, $t = 20.2$ s - Vertical displacement (height), $h = 2000$ m 3. **Calculate horizontal displacement:** Since horizontal velocity is constant, $$x = v_x \times t = 100 \times 20.2 = 2020 \text{ m}$$ 4. **Calculate vertical displacement:** The box falls 2000 m vertically downward. 5. **Calculate the angle of displacement:** The displacement vector has components $x = 2020$ m (horizontal) and $y = 2000$ m (vertical downward). The angle $\theta$ between the horizontal and the displacement vector is given by $$\theta = \tan^{-1}\left(\frac{\text{vertical displacement}}{\text{horizontal displacement}}\right) = \tan^{-1}\left(\frac{2000}{2020}\right)$$ 6. **Evaluate the angle:** $$\theta = \tan^{-1}(0.9901) \approx 44.5^\circ$$ 7. **Final answer:** The angle between the horizontal and the displacement of the box is approximately **44.5 degrees** to 3 significant figures.