Q 11-02-111JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
A balloon was moving upwards with a uniform velocity of $10$ m s$^{-1}$. 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 object strikes the ground was around: (takes the value of $g$ as $10$ m s$^{-2}$)
Answer: (C) $125$ m
The object starts with the balloon's velocity, $10$ m s$^{-1}$ upward. Taking up as positive:
$$-75 = 10t - 5t^2 \Rightarrow t^2 - 2t - 15 = 0 \Rightarrow t = 5\ \text{s}$$
In $5$ s the balloon rises $10\times5 = 50$ m, so its height is $75 + 50 = 125$ m.
Solution by Sreeraj P, M.Sc Physics