A stone is thrown vertically upward with a speed of 28 m/s ....

Using the third equation of motion, $v^2 = u^2 - 2gH$.

At maximum height, $v = 0$. Thus, $H = \dfrac{u^2}{2g}$.

Substituting $u = 28 \text{ m/s}$ and $g = 9.8 \text{ m/s}^2$:

$H = \dfrac{28^2}{2 \times 9.8} = \dfrac{784}{19.6} = 40 \text{ m}$.

Let $T$ be the time taken to reach maximum height. At maximum height, $v = 0$, so $0 = u - gT \Rightarrow T = \dfrac{u}{g}$.

The velocity one second before reaching maximum height is at time $t = T - 1$.

Using $v' = u - gt$:

$v' = u - g(T - 1) = u - g\left(\dfrac{u}{g} - 1\right) = u - u + g = g = 9.8 \text{ m/s}$.

Since $v'$ depends only on $g$ and is independent of the initial velocity $u$, this velocity will not change if the initial speed is $40 \text{ m/s}$ or $80 \text{ m/s}$.