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
Concept. Find the speed at the top of the window from the crossing time, then the drop needed to reach that speed.
Step 1. Speed at the top of the window.
$$2.2 = 0.2v + \frac12(10)(0.2)^2 = 0.2v + 0.2 \;\Rightarrow\; v = 10\ \text{m/s}$$
Step 2. Height above the window.
$$h = \frac{v^2}{2g} = \frac{100}{20} = 5\ \text{m}$$
Answer. $5$ m (Option D).
Solution by Sreeraj P, M.Sc Physics