Answer to Question #311130 in Physics for 4rme

Question

A projectile has an initial launch speed of 18.0 m/s. If it is desired that it strikes a target 31.0 m away at the same elevation, what should be the projection angle?

Answer 1

The range

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

Hence

$$\sin2\theta=gR/v_0^2\\ =9.81*31.0/18.0^2=0.939$$

$$\theta=34.9^\circ$$