The helicopter rises from rest on the ground vertically upwards with a constant acceleration g/8 . A food packet is dropped from the helicopter when it has risen to a height h . Show that the time taken by the packet to reach the ground is 2{h/g}^1/2​

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Answer:

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]:

1) t = 3.4 √(h / g)

2) t = √(2h / 3g)

3) t = (2 / 3) √(h / g)

4) t = 1.8 √(h / g)

Answer: (1)

JEE Main 2020 Physics Paper Shift 1 Sept 5 Question 6

VB2 = 02 + 2gh

VB = √2gh

– h = (VB) t – (1 / 2) gt2

– h = √2gh * t – (1 / 2) gt2

gt2 – 2 √2gh * t – 2h = 0

Explanation:

Explanation:

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