Skip to content

JEE MAIN 2026 · Kinematics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 29

Updated
Single correct+4 / -1Answer matches NTA’s final key

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=10g = 10 m/s²)

Choose one

Solution

Solution

At release the stone shares the balloon's velocity: 1010 m/s upward. Taking up as positive, it reaches the ground when

−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}.

In those 5 s the balloon rises a further 10×5=5010 \times 5 = 50 m, so it is at 75+50=12575 + 50 = 125 m.

Answer: 125 m.

Answer checked against the official answer key. The solution is our own working.

Keep going

A gas balloon is going up with a constant… | SchoolToCareer