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]
Pick an option to check your answer.
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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