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) $20\ \text{m s}^{-1}$
The stone is dropped, so $u = 0$. Using $v^2 = u^2 + 2gh$:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics