Q 11-02-094JEE MainJEE Main 2022 (26 Jun, Shift 1)Easy
A ball of mass $0.5\ \text{kg}$ is dropped from the height of $10\ \text{m}$. The height at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity is ______ m. [Use $g = 10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 5
We need $v = 10\ \text{m s}^{-1}$. Distance fallen: $\dfrac{v^2}{2g} = \dfrac{100}{20} = 5\ \text{m}$, so the height above the ground is $10 - 5 = 5\ \text{m}$.
Solution by Sreeraj P, M.Sc Physics