1. **State the problem:** Find the vertex of the quadratic function $h(t) = -16t^2 + 40t$. 2. **Formula and rules:** The vertex of a quadratic function $h(t) = at^2 + bt + c$ is given by the point $$\left(-\frac{b}{2a}, h\left(-\frac{b}{2a}\right)\right).$$ Here, $a = -16$, $b = 40$, and $c = 0$. 3. **Calculate the $t$-coordinate of the vertex:** $$t = -\frac{b}{2a} = -\frac{40}{2 \times -16} = -\frac{40}{-32} = \frac{40}{32}.$$ 4. **Simplify the fraction:** $$\frac{40}{32} = \frac{\cancel{8}5}{\cancel{8}4} = \frac{5}{4} = 1.25.$$ 5. **Calculate the $h(t)$ value at $t=1.25$:** $$h(1.25) = -16(1.25)^2 + 40(1.25) = -16(1.5625) + 50 = -25 + 50 = 25.$$ 6. **Final answer:** The vertex of the function is at $$\boxed{\left(1.25, 25\right)}.$$