JEE Mains · Physics · STD 11 - 2. motion in straight line
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]
- A \(t =\sqrt{\frac{2 h }{3 g }}\)
- B \(t=1.8 \sqrt{\frac{h}{g}}\)
- C \(t=3.4 \sqrt{\left(\frac{h}{g}\right)}\)
- D \(t=\frac{2}{3} \sqrt{\left(\frac{h}{g}\right)}\)
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…
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