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A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]

1. t = √(2h/3g)
2. t = 1.8 √(h/g)
3. t = 3.4 √(h/g)
4. t = 2/3 √(h/g)

Correct answer: t = 3.4 √(h/g)

Solution

At height h the helicopter speed is v = sqrt(2gh) upward; the packet then has u=+sqrt(2gh), y0=h, a=-g. Solving h + sqrt(2gh) t - (1/2) g t^2 = 0 gives t = (2 + sqrt2) sqrt(h/g) ~ 3.4 sqrt(h/g).

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