Let the total time of fall be $T$ seconds.

Since the ball is dropped from rest from a height of $100\,\text{m}$, 100 = \frac{1}{2}gT^2 \quad \Rightarrow \quad gT^2 = 200 \tag{1}

Distance covered in the last $\frac{1}{2}$ s

Distance fallen in time $t$ from rest is $s(t)=\frac{1}{2}gt^2$

Therefore, distance covered in the last $0.5$ s is $s(T)-s(T-0.5)=19$

$\frac{1}{2}gT^2-\frac{1}{2}g(T-0.5)^2=19$

Simplify the above equation

$\frac{1}{2}g\left[T^2-(T-0.5)^2\right]=19$

Now, $T^2-(T-0.5)^2 = T^2-(T^2-T+0.25)=T-0.25$

So, $\frac{1}{2}g(T-0.25)=19$ g(T-0.25)=38 \tag{2}

Use equation (1) to solve

From (1), $g=\frac{200}{T^2}$

Substitute into (2): $\frac{200}{T^2}(T-0.25)=38$

$200T-50=38T^2$

$38T^2-200T+50=0$

Divide by $2$: $19T^2-100T+25=0$

Solve the quadratic

$T=\frac{100\pm\sqrt{10000-4\cdot 19\cdot 25}}{2\cdot 19}$ $T=\frac{100\pm\sqrt{8100}}{38}$ $T=\frac{100\pm 90}{38}$

So, $T=5\,\text{s} \quad \text{or} \quad T=\frac{10}{38}=\frac{5}{19}\,\text{s}$

Since the total time must be greater than $0.5$ s, we take $T=5\,\text{s}$

Find $g$

Using (1): $g=\frac{200}{T^2}=\frac{200}{25}=8\,\text{m/s}^2$

Final answer

$\boxed{8}$