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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2020
  • Hard
  • MCQ

A helicopter reises 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]

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Answer & Solution

Correct Answer

(C) \(t=3.4 \sqrt{\left(\frac{h}{g}\right)}\)

Step-by-step Solution

Detailed explanation

\(t_{0}=\sqrt{\frac{2 h}{g}}\)

\(\Rightarrow \quad u=\sqrt{\frac{2 h}{g}} \times g=\sqrt{2 g h}\)

\(\therefore \quad t_{1}=\)time to reach top \(=\frac{u}{g}=\sqrt{\frac{2 h}{g}}\)

\(\therefore \quad H=h+h^{\prime}=2 h\)

\(t_{2}=\) time of fall \(=\sqrt{\frac{2 \times(2 h)}{g}}\)

\(\quad=2 \sqrt{\frac{h}{g}}\)

Total time \(=t_{1}+t_{2}\)

\(\quad=(2+\sqrt{2}) \sqrt{\frac{h}{g}}\)

\(\quad=3.4 \sqrt{\frac{h}{g}}\)

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