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 : (use acceleration due to gravity g = 10 m/s2 and neglect air resistance)
- 7.2 km
- 2 5 km
- 2 2 km
- 4 km
Answer
(3) 2 2 km
Solution
Horizontal velocity of the object = 360 km/h = 360 518 = 100 m/s
Horizontal displacement in 20 s: x = 100 20 = 2000 m = 2 km
Vertical displacement: y = 12gt2 = 12 10 202 = 2000 m = 2 km (equal to the altitude ✓)
displacement = x2 + y2 = 22 + 22 = 2 2 km
Answer: (3)