Q 11-02-031NEETJEE MainMedium
A ball is dropped from the roof of a building. It takes $0.2$ s to fall past a window $2.2$ m tall. How far above the top of the window is the roof? ($g = 10\ \text{m/s}^2$)
Answer: (D) $5$ m
Let $v$ be the speed at the top edge of the window. Across the window:
$$2.2 = v(0.2) + \frac{1}{2}(10)(0.2)^2 = 0.2v + 0.2 \;\Rightarrow\; v = 10\ \text{m/s}$$
The ball fell from rest to reach this speed:
$$h = \frac{v^2}{2g} = \frac{100}{20} = 5\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics