Given the function

$$
\(y=8 t-\frac{t^2}{2}-24\)
$$

we want to find its maximum $y$-value.
Step 1: Rewrite the function in a clearer form

$$
\(y=8 t-\frac{1}{2} t^2-24\)
$$


This is a quadratic function of the form:

$$
\(y=-\frac{1}{2} t^2+8 t-24\)
$$


Here:
- $\(a=-\frac{1}{2}\)$ (coefficient of $\(t^2\)$ ),
- $\(b=8\)$ (coefficient of $t$ ),
- $\(c=-24\)$ (constant term).

Since $a<0$, the parabola opens downward and the function has a maximum value at its vertex.
Step 2: Find the $t$-value of the vertex
The vertex $t$-coordinate is given by:

$$
\(t=-\frac{b}{2 a}=-\frac{8}{2 \times\left(-\frac{1}{2}\right)}=-\frac{8}{-1}=8\)
$$


Step 3: Calculate the maximum $y$-value by substituting $t=8$ into the function

$$
\(y=8(8)-\frac{1}{2}(8)^2-24=64-\frac{1}{2} \times 64-24=64-32-24=8 \).
$$


Final answer:
The maximum value of $y$ is 8.