Initial velocity u = 100 m/s

The vertical component of velocity at the highest point is 0.

$$t_{max} = \frac{u \sin \theta}{g}$$

$$t_{max} = \frac{100 \, \text{m/s} \times \sin(30^\circ)}{10 \, \text{m/s}^2}$$ $$t_{max} = \frac{100 \, \text{m/s} \times 0.5}{10 \, \text{m/s}^2}$$ $$t_{max} = \frac{50 \, \text{m/s}}{10 \, \text{m/s}^2}$$ $$t_{max} = 5 \, \text{s}$$

The time to reach the highest point is $\boxed{\text{5 s}}$.

$$H_{max} = \frac{(100 \, \text{m/s})^2 \times (\sin(30^\circ))^2}{2 \times 10 \, \text{m/s}^2}$$ $$H_{max} = \frac{10000 \, \text{m}^2/\text{s}^2 \times (0.5)^2}{20 \, \text{m/s}^2}$$ $$H_{max} = \frac{10000 \, \text{m}^2/\text{s}^2 \times 0.25}{20 \, \text{m/s}^2}$$ $$H_{max} = \frac{2500 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2}$$ $$H_{max} = 125 \, \text{m}$$

The maximum height reached is $\boxed{\text{125 m}}$.

$$R = \frac{(100 \, \text{m/s})^2 \times \sin(2 \times 30^\circ)}{10 \, \text{m/s}^2}$$ $$R = \frac{10000 \, \text{m}^2/\text{s}^2 \times \sin(60^\circ)}{10 \, \text{m/s}^2}$$ $$R = \frac{10000 \, \text{m}^2/\text{s}^2 \times \frac{\sqrt{3}}{2}}{10 \, \text{m/s}^2}$$ $$R = \frac{10000 \, \text{m}^2/\text{s}^2 \times 0.866025}{10 \, \text{m/s}^2}$$ $$R = 1000 \, \text{m} \times 0.866025$$ $$R \approx 866.025 \, \text{m}$$

The horizontal range of the ball is $\boxed{\text{866 m}}$.