A helicopter flying horizontally with a speed of 360~km/h$360~km/h$ at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is: (use acceleration due to gravity g=10~m/s^2$g=10~m/s^{2}$ and neglect air resistance) [cite: 157, 158, 159, 160]

### Related Formula x = u cdot t$x = u \cdot t$ [cite: 763] H = frac12gt^2$$H = \frac{1}{2}gt^2$$ [cite: 764] D = sqrtx^2 + H^2$$D = \sqrt{x^2 + H^2}$$ [cite: 774] ### Core Logic First, convert the horizontal velocity component to metric SI units: [cite: 157, 763] u = 360 times frac518 = 100\ textm/s$$u = 360 \times \frac{5}{18} = 100\ \text{m/s}$$ [cite: 157, 763] Calculate the horizontal range distance x$x$ covered over t = 20\ texts$t = 20\ \text{s}$: [cite: 158, 763] x = 100 times 20 = 2000\ textm = 2\ textkm$$x = 100 \times 20 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 763] The vertical displacement height H$H$ is explicitly given as 2\ textkm = 2000\ textm$2\ \text{km} = 2000\ \text{m}$[cite: 157, 769]. Let's confirm with free-fall height calculation matching the solution template: [cite: 764] H = frac12 times 10 times (20)^2 = 5 times 400 = 2000\ textm = 2\ textkm$$H = \frac{1}{2} \times 10 \times (20)^2 = 5 \times 400 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 769] Now find the net spatial vector displacement D$D$ from the release coordinates: [cite: 774] D = sqrtx^2 + H^2 = sqrt2^2 + 2^2 = sqrt8 = 2sqrt2\ textkm$$D = \sqrt{x^2 + H^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\ \text{km}$$ [cite: 774] ### Pattern Recognition Be careful with the wording: the question asks for the displacement from the *release coordinate position* [cite: 159], which is the hypotenuse vector sqrtx^2 + H^2$\sqrt{x^2 + H^2}$[cite: 774]. Do not mistake it for the horizontal range distance alone. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane