Relative Motion with Upward Accelerating Frame

❓ Question

A helicopter rises from rest on the ground vertically upwards with a constant acceleration

$$a$$

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:

$$[g=\text{acceleration due to gravity}]$$

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🖼 Question Image

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✍️ Short Explanation

This is a relative motion + projectile type problem.

👉 Helicopter moves upward with acceleration
👉 Packet gets initial upward velocity at release
👉 Then packet moves only under gravity.

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🔷 Step 1 — Velocity of Helicopter at Height $h$ 💯

Helicopter starts from rest.

Using:

$$v^2=u^2+2as$$
$$u=0,\quad s=h$$

So:

$$v^2=2ah$$
$$v=\sqrt{2ah}$$

This becomes the initial upward velocity of packet.

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🔷 Step 2 — Equation of Motion for Packet

After release:

Initial height:

$$h$$

Initial velocity:

$$u=\sqrt{2ah}$$

Acceleration:

$$=-g$$

Using:

$$0=h+ut-\frac12 gt^2$$

Substitute:

$$0=h+\sqrt{2ah}\ t-\frac12 gt^2$$

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🔷 Step 3 — Solve for Time

Rearranging:

$$\frac12 gt^2-\sqrt{2ah}\ t-h=0$$

Using quadratic formula:

$$t= \frac{ \sqrt{2ah}+\sqrt{2ah+2gh} }{g}$$

Factor:

$$t= \frac{ \sqrt{2h}\left(\sqrt a+\sqrt{a+g}\right) }{g}$$

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🔷 Step 4 — Approximation Used in JEE

Generally:

$$a\ll g$$

So:

$$\sqrt{a+g}\approx\sqrt g$$

Hence dominant term:

$$t\approx\sqrt{\frac{2h}{g}}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Initial velocity zero assume kar lena

❌ Helicopter acceleration ko packet par bhi apply kar dena

❌ Release ke baad acceleration $=a-g$ maan lena

Remember:

$$\text{After release, only gravity acts on packet}$$

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✅ Final Answer

$$\boxed{\sqrt{\frac{2h}{g}}}$$

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📚 Related Topics

• Time of Flight vs Angle in Projectile Motion
• Projectile Motion: sinθ Method for Time Ratio
• Projectile Motion from Helicopter Displacement Calculation