The maximum height $s$ reached by a ball thrown vertically upward with initial velocity $u$ (taking magnitude of gravity as $g$) is given by:

1. At maximum height the final velocity $v = 0$ and acceleration $a = -g$. 2. From $v^2 = u^2 + 2as$: $0 = u^2 + 2(-g)s$. 3. Rearranging, $2gs = u^2$, so $s = \dfrac{u^2}{2g}$. 4. Dropping the square on $u$ or omitting the factor 2 gives the wrong options. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.22_