A body is projected upwards with an initial velocity of 30 m/s. If the horizontal range of projection is 75 m, at what angle with the horizontal is the ball projected? [g = 10 ms-2]

The horizontal range $R$ of a projectile launched with initial velocity $u$ at an angle $\theta$ to the horizontal is given by:

$$R = \frac{u^2 \sin(2\theta)}{g}$$

We are given $R = 75 \text{ m}$, $u = 30 \text{ ms}^{-1}$, and $g = 10 \text{ ms}^{-2}$.

$$75 = \frac{(30)^2 \sin(2\theta)}{10}$$

$$75 = \frac{900 \sin(2\theta)}{10}$$ $$75 = 90 \sin(2\theta)$$

$$\sin(2\theta) = \frac{75}{90}$$ $$\sin(2\theta) = \frac{5}{6}$$ $$\sin(2\theta) \approx 0.8333$$

$$2\theta = \arcsin\left(\frac{5}{6}\right)$$ $$2\theta \approx 56.44^\circ$$

$$\theta = \frac{56.44^\circ}{2}$$ $$\theta \approx 28.22^\circ$$

The angle with the horizontal at which the ball is projected is approximately $28.2^\circ$.