A helicopter flying horizontally with a speed of 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 :
❓ Question A helicopter flying horizontally with a speed of 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 seconds after it is dropped. Find the displacement of point O from the position of the helicopter where the object was released. 🖼️ Question Image ✍️ Short Solution Let’s calculate step-by-step: Step 1: Horizontal Distance (Range) x = u x × t = 100 × 20 = 2000 m Step 2: Vertical Motion Check Vertical displacement under gravity: y = 1 2 g t 2 = 1 2 × 9.8 × 20 2 = 1960 m which is approximately equal to the helicopter’s altitude (2000 m). ✅ Hence, the time given (20 s) is consistent. Step 3: Resultant Displacement The object moves 2000 m horizontally and 2000 m vertically downward , so total displacement: s = x 2 + y 2 = ( 2000 ) 2 + ( 2000 ) 2 = 2000 2 m s = \sqrt{x^2 + y^2} = \sqrt{(2000)^2 + (2000)^2} = 2000\sqrt{2}\,\text{m} s = 2.83 km (approximately) 🧮 Image Solution ...