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 at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]

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(3) t = 3.4√(/hg). Helicopter from rest with acceleration g: at height h its speed is v = √2gh, upward. The packet leaves with this upward velocity and then falls freely. Upward positive: -h = √2gh t - 1/2gt². 1/2gt² - √2gh t - h = 0 ⇒ t = (√2gh + √2gh + 2gh)/g = (√2gh + 2√gh)/g. t = (√2 + 2)√/hg = 3.4√/hg.

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 at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]

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