Q 11-02-030NEETJEE MainHard
A ball projected vertically upward passes a point A with speed $\dfrac{u}{2}$ and a point B, $9$ m higher, with speed $\dfrac{u}{4}$, where $u$ is its speed of projection. The maximum height reached by the ball above the point of projection is
Answer: (C) $48$ m
Concept. Apply $v^2 = u^2 - 2gh$ between A and B to find $u^2$ in terms of $g$; then $H = \dfrac{u^2}{2g}$.
Step 1. Between A and B.
$$\left(\frac{u}{2}\right)^2 - \left(\frac{u}{4}\right)^2 = 2g(9) \;\Rightarrow\; \frac{3u^2}{16} = 18g \;\Rightarrow\; u^2 = 96g$$
Step 2. Maximum height. $H = \dfrac{96g}{2g} = 48$ m.
Answer. $48$ m (Option C).
Solution by Sreeraj P, M.Sc Physics