How does the horizontal component of a vector change when the angle with the horizontal increases from 30° to 60°?

The horizontal component ($V_x$) of a vector with magnitude $V$ at an angle $\theta$ with the horizontal is given by:

$$V_x = V \cos \theta$$

$$V_{x, \text{initial}} = V \cos 30^\circ$$

We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

$$V_{x, \text{initial}} = V \frac{\sqrt{3}}{2}$$

$$V_{x, \text{final}} = V \cos 60^\circ$$

We know that $\cos 60^\circ = \frac{1}{2}$.

$$V_{x, \text{final}} = V \frac{1}{2}$$

To find how the component changes, we can express the final value in terms of the initial value:

$$\frac{V_{x, \text{final}}}{V_{x, \text{initial}}} = \frac{V \frac{1}{2}}{V \frac{\sqrt{3}}{2}}$$ $$\frac{V_{x, \text{final}}}{V_{x, \text{initial}}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$ $$\frac{V_{x, \text{final}}}{V_{x, \text{initial}}} = \frac{1}{2} \times \frac{2}{\sqrt{3}}$$ $$\frac{V_{x, \text{final}}}{V_{x, \text{initial}}} = \frac{1}{\sqrt{3}}$$

So, $V_{x, \text{final}} = \frac{1}{\sqrt{3}} V_{x, \text{initial}}$.

The horizontal component becomes $\frac{1}{\sqrt{3}}$ times its original value.

The final answer is $\text{D}$.