Balloon Motion and Falling Object Height Problem

❓ Question

A balloon was moving upwards with a uniform velocity of:

$$10\text{ m/s}$$

An object of finite mass is dropped from the balloon when it was at a height of:

$$75\text{ m}$$

from the ground level.

Find the height of the balloon from the ground when the object strikes the ground.

Take:

$$g=10\text{ m/s}^2$$

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🖼 Question Image

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✍️ Short Explanation

When object is released from moving balloon:

👉 Initial velocity of object remains same as balloon velocity
👉 Object first moves upward
👉 Then comes down under gravity.

Meanwhile balloon keeps moving upward uniformly.

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🔷 Step 1 — Initial Conditions 💯

Height from ground:

$$u_h=75\text{ m}$$

Initial upward velocity:

$$u=10\text{ m/s}$$

Acceleration:

$$a=-10\text{ m/s}^2$$

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🔷 Step 2 — Time Taken by Object to Reach Ground

Using:

$$s=ut+\frac12 at^2$$

Taking upward positive:

$$-75=10t-\frac12(10)t^2$$
$$-75=10t-5t^2$$
$$5t^2-10t-75=0$$
$$t^2-2t-15=0$$
$$(t-5)(t+3)=0$$
$$t=5\text{ s}$$

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🔷 Step 3 — Motion of Balloon

Balloon keeps moving upward with constant velocity:

$$v=10\text{ m/s}$$

Distance moved in 5 s:

$$s=vt$$
$$s=10\times5$$
$$s=50\text{ m}$$

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🔷 Step 4 — Final Height of Balloon

Initial height:

$$75\text{ m}$$

Extra upward distance:

$$50\text{ m}$$

So final height:

$$75+50=125\text{ m}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Initial velocity zero assume kar lena

❌ Balloon ko acceleration dena

❌ Object aur balloon ka same motion assume kar lena

Remember:

$$\text{Released object keeps balloon's velocity initially}$$

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✅ Final Answer

$$\boxed{125\text{ m}}$$

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