Q 11-02-013NEETAIPMT 2011Easy
A boy standing at the top of a tower of $20$ m height drops a stone. Assuming $g = 10\ \text{m s}^{-2}$, the velocity with which it hits the ground is
Answer: (A) $20\ \text{m s}^{-1}$
Concept. A dropped body ($u = 0$) reaches $v = \sqrt{2gh}$.
Step 1. Substitute.
$$v = \sqrt{2 \times 10 \times 20} = 20\ \text{m s}^{-1}$$
Answer. $20$ m s$^{-1}$ (Option A).
Solution by Sreeraj P, M.Sc Physics