Question:
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]
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]
[\(g\) is the acceleration due to gravity]
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When dropped, the packet has the helicopter velocity \(\sqrt{2gh}\) upward. Solve \(-h=ut-\frac{1}{2}gt^{2}\) for \(t\).
Updated On: Nov 5, 2026
- \(t=3.4\sqrt{\left(\frac{h}{g}\right)}\)
- \(t=1.8\sqrt{\left(\frac{h}{g}\right)}\)
- \(t=\sqrt{\frac{2h}{3g}}\)
- \(t=\frac{2}{3}\sqrt{\left(\frac{h}{g}\right)}\)
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The Correct Option is A
Solution and Explanation
Helicopter speed at height \(h\): \(u^{2}=2gh\Rightarrow u=\sqrt{2gh}\) (upward).
The packet starts with this upward velocity and then has acceleration \(-g\). Taking up as positive and the drop point as origin, its displacement on reaching the ground is \(-h\):
\(-h=ut-\frac{1}{2}gt^{2}\Rightarrow gt^{2}-2ut-2h=0\).
\(t=\frac{2u+\sqrt{4u^{2}+8gh}}{2g}=\frac{u+\sqrt{u^{2}+2gh}}{g}=\frac{\sqrt{2gh}+\sqrt{4gh}}{g}\).
\(t=\left(\sqrt{2}+2\right)\sqrt{\frac{h}{g}}=3.41\sqrt{\frac{h}{g}}\approx3.4\sqrt{\frac{h}{g}}\).
Answer: (1) \(t=3.4\sqrt{\left(\frac{h}{g}\right)}\)
The packet starts with this upward velocity and then has acceleration \(-g\). Taking up as positive and the drop point as origin, its displacement on reaching the ground is \(-h\):
\(-h=ut-\frac{1}{2}gt^{2}\Rightarrow gt^{2}-2ut-2h=0\).
\(t=\frac{2u+\sqrt{4u^{2}+8gh}}{2g}=\frac{u+\sqrt{u^{2}+2gh}}{g}=\frac{\sqrt{2gh}+\sqrt{4gh}}{g}\).
\(t=\left(\sqrt{2}+2\right)\sqrt{\frac{h}{g}}=3.41\sqrt{\frac{h}{g}}\approx3.4\sqrt{\frac{h}{g}}\).
Answer: (1) \(t=3.4\sqrt{\left(\frac{h}{g}\right)}\)
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