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JEE Main 2026 (8 Apr, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2026 (8 Apr, Shift 2)

Q8JEE Main 2026 · 8 APR · SHIFT 2Medium

Motion Under Gravity

A gas balloon is going up with a constant velocity of 10 m/s10\text{ 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/s2g = 10\text{ m/s}^2)

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Answer: (d)

  1. Stone leaves with the balloon's velocity, 10 m/s10\text{ m/s} upward, at height 75 m.
  2. −75=10t−5t2⇒t2−2t−15=0⇒t=5 s-75 = 10t - 5t^2 \Rightarrow t^2 - 2t - 15 = 0 \Rightarrow t = 5\text{ s}
  3. Balloon height =75+10×5=125 m= 75 + 10 \times 5 = 125\text{ m}
  4. So the answer is (d) 125.
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