Question

Compute the horizontal range, on level ground, for a projectile with initial speed 35.0m/s and initial inclination 75.0⁰ from the horizontal.

Answer 1

The horizontal range is given by

$$R=\frac{v_0^2\sin2\theta}{g}$$

$$R=\frac{35.0^2\sin(2*75.0^\circ)}{9.8}=62.5\:\rm m$$