🎈 Balloon Drops Object – Vertical Motion with Initial Velocity | JEE Physics | Doubtify JEE

🎯 Question:

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground, when the object strikes the ground, was around:

(Take g = 10 m/s²)

📌 Why This Question is Important:

This is a classic JEE Physics conceptual question from the chapter Motion in a Straight Line. It tests your understanding of:

• Relative motion and initial velocity

• Vertical motion equations

• Kinematics with constant acceleration

Many students make mistakes by assuming the object is dropped with zero velocity, forgetting that it inherits the balloon's upward velocity at the moment of release. Mastering this kind of problem clears your concepts on vertical motion and helps you solve both NEET and JEE questions accurately.

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💡 Concept Used:

The object, though dropped (i.e., no extra force), already has an initial upward velocity (u = 10 m/s) due to the motion of the balloon. After being dropped, it undergoes uniformly accelerated motion under gravity (a = –g = –10 m/s²).

We use the kinematic equation to calculate time of flight of the object:

$$h=ut+\frac{1}{2}(-g)t^2$$

Putting values:

$$75=10t-5t^2\Rightarrow 5t^2-10t+75=0$$

Solving this quadratic:

$$t=5\text{seconds}$$

So, the object takes 5 seconds to hit the ground. Since the balloon is still moving upwards at 10 m/s, in 5 seconds it rises:

$$\text{Height risen}=v\cdot t=10\times 5=50\text{m}$$

Hence, final height of balloon = 75 + 50 = 125 meters​

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🧠 Solution (Image):

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🎥 Video Solution:

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📌 Remember: The key to mastering kinematics is understanding motion before and after the action.
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