Q 11-02-046JEE MainJEE Main 2026 (5 Apr, Shift 1)Easy
From $18$ m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m. (Take $g = 10\ \text{m/s}^2$ and neglect the air resistance)
Numerical value type. Enter your answer.
Answer: 13
Concept. For a dropped body $v^2=2gs$, where $s$ is the distance fallen. "Velocity equal to $g$ numerically" means $v=10$ m/s.
Step 1. Distance fallen.
$$s=\frac{v^2}{2g}=\frac{100}{20}=5\ \text{m}$$
Step 2. Height above the ground. $18-5=13$ m.
Answer. $13$.
Solution by Sreeraj P, M.Sc Physics
