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
Between A and B (rising $9$ m):
$$\left(\frac{u}{2}\right)^2 - \left(\frac{u}{4}\right)^2 = 2g(9) \;\Rightarrow\; \frac{3u^2}{16} = 18g \;\Rightarrow\; u^2 = 96g$$
Maximum height above the point of projection:
$$H = \frac{u^2}{2g} = \frac{96g}{2g} = 48\ \text{m}$$
(The value of $g$ cancels, so the answer does not depend on it.)
Solution by Sreeraj P, M.Sc Physics