The conical pendulum consists of a mass $m$ rotating in a horizontal circle of radius $r$ at the end of a string of length $L$. The string makes an angle $\theta$ with the vertical. The forces acting on the mass are the tension $T$ in the string and the gravitational force $mg$.

We apply Newton's second law by resolving the forces into vertical and horizontal components. The vertical components of the forces must balance, as there is no vertical acceleration:

$$T \cos\theta - mg = 0 \implies T \cos\theta = mg \quad (1)$$

The horizontal component of the tension provides the necessary centripetal force, $F_c = ma_c$, for the circular motion:

$$T \sin\theta = ma_c \quad (2)$$

The radius of the circular path is $r = L \sin\theta$. The angular velocity is $\omega = 2\pi/P$, where $P$ is the period. The centripetal acceleration is $a_c = \omega^2 r$. Substituting these into equation (2):

$$T \sin\theta = m \left(\frac{2\pi}{P}\right)^2 (L \sin\theta)$$

To find $g$, we can divide the horizontal force equation by the vertical force equation to eliminate the tension $T$ and mass $m$:

$$\frac{T \sin\theta}{T \cos\theta} = \frac{m \left(\frac{2\pi}{P}\right)^2 L \sin\theta}{mg}$$ $$\tan\theta = \frac{\left(\frac{2\pi}{P}\right)^2 L \sin\theta}{g}$$

Now, we solve for the acceleration due to gravity, $g$:

$$g = \frac{\left(\frac{2\pi}{P}\right)^2 L \sin\theta}{\tan\theta} = \left(\frac{2\pi}{P}\right)^2 L \cos\theta$$

Finally, we substitute the given values: $L = 0.28$ m, $P = 1.00$ s, and $\theta = 30^\circ$.

$$g = \left(\frac{2\pi}{1.00 \text{ s}}\right)^2 (0.28 \text{ m}) \cos(30^\circ)$$ $$g = (4\pi^2)(0.28)\left(\frac{\sqrt{3}}{2}\right)$$ $$g \approx (39.478)(0.28)(0.8660) \approx 9.57 \text{ m/s}^2$$