A gas balloon is going up with a constant velocity of $10\ \text{m/s}$. 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 $g = 10\ \text{m/s}^2$)
Answer: (D) $125$
Concept. A stone released from a moving balloon starts with the balloon's velocity ($10$ m/s upward), not from rest. After release only gravity acts on it, so $s=ut-\tfrac12gt^2$ applies with that $u$. Meanwhile the balloon keeps rising at constant speed.
Step 1. Equation for the stone. Take upward as positive, origin at the release point. The ground is at $s=-75$ m:
$$-75=10t-\tfrac12(10)t^2=10t-5t^2$$
Step 2. Solve the quadratic.
$$5t^2-10t-75=0\ \Rightarrow\ t^2-2t-15=0\ \Rightarrow\ (t-5)(t+3)=0\ \Rightarrow\ t=5\ \text{s}$$
(The negative root has no meaning here.)
Step 3. Height of the balloon then. It rises $10\times5=50$ m more:
$$H=75+50=125\ \text{m}$$
Answer. $125$ m (Option D). Taking the stone as dropped from rest gives $t=\sqrt{15}$ s and a wrong height.
Solution by Sreeraj P, M.Sc Physics
