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JEE Main 2026 · Physics
Kinematics
A gas balloon is going up with a constant velocity of 10 m/s. When this balloon reached a height of 75 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is _______ m. (Take g = 10 m/s²)
- A85
- B150
- C129
- D125
Rough work
- Crossed out: 75 = 5t², so t = 3.9 s
- Crossed out: 75 + 10 × 3.9 = 114
- y = 75 + 10t − 5t²
- 5t² − 10t − 75 = 0, t = 5
- Balloon: 75 + 10 × 5 = 125
Solved
Polaris, your AI tutor
Polaris, your AI tutor: I'm right here. Tell me what you've tried, or where it stopped making sense.
You: I got 114, but that's not one of the options.
You: It falls 75 m from rest, so I used s = ½gt².
You: No, it was moving with the balloon. But how do I write its height now?
You: So it starts at 10 m/s, going up.
Questions about NEET
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