Here are the calculations for the projectile motion problem. Given: Initial velocity $u = 100 \, \text{m/s}$ Angle of projection $\theta = 30^\circ$ Acceleration due to gravity $g = 10 \, \text{m/s}^2$ a) The time at which the ball reaches the highest point ($t_{max}$) Step 1: Use the formula for the time to reach maximum height. The vertical component of velocity at the highest point is $0$. $$ t_{max} = \frac{u \sin \theta}{g} $$ Step 2: Substitute the given values into the formula. $$ t_{max} = \frac{100 \, \text{m/s} \times \sin(30^\circ)}{10 \, \text{m/s}^2} $$ $$ t_{max} = \frac{100 \, \text{m/s} \times 0.5}{10 \, \text{m/s}^2} $$ $$ t_{max} = \frac{50 \, \text{m/s}}{10 \, \text{m/s}^2} $$ $$ t_{max} = 5 \, \text{s} $$ The time to reach the highest point is $\boxed{\text{5 s}}$. b) The maximum height reached ($H_{max}$) Step 1: Use the formula for the maximum height. $$ H_{max} = \frac{u^2 \sin^2 \theta}{2g} $$ Step 2: Substitute the given values into the formula. $$ H_{max} = \frac{(100 \, \text{m/s})^2 \times (\sin(30^\circ))^2}{2 \times 10 \, \text{m/s}^2} $$ $$ H_{max} = \frac{10000 \, \text{m}^2/\text{s}^2 \times (0.5)^2}{20 \, \text{m/s}^2} $$ $$ H_{max} = \frac{10000 \, \text{m}^2/\text{s}^2 \times 0.25}{20 \, \text{m/s}^2} $$ $$ H_{max} = \frac{2500 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} $$ $$ H_{max} = 125 \, \text{m} $$ The maximum height reached is $\boxed{\text{125 m}}$. c) The horizontal range of the ball ($R$) Step 1: Use the formula for the horizontal range. $$ R = \frac{u^2 \sin(2\theta)}{g} $$ Step 2: Substitute the given values into the formula. $$ R = \frac{(100 \, \text{m/s})^2 \times \sin(2 \times 30^\circ)}{10 \, \text{m/s}^2} $$ $$ R = \frac{10000 \, \text{m}^2/\text{s}^2 \times \sin(60^\circ)}{10 \, \text{m/s}^2} $$ $$ R = \frac{10000 \, \text{m}^2/\text{s}^2 \times \frac{\sqrt{3}}{2}}{10 \, \text{m/s}^2} $$ $$ R = \frac{10000 \, \text{m}^2/\text{s}^2 \times 0.866025}{10 \, \text{m/s}^2} $$ $$ R = 1000 \, \text{m} \times 0.866025 $$ $$ R \approx 866.025 \, \text{m} $$ The horizontal range of the ball is $\boxed{\text{866 m}}$.