Answer to Question #176696 in Physics for Francine

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

We can find the projection angle from the definition of the range:

$$R=\dfrac{v_0^2sin2\theta}{g},$$ $$sin2\theta=\dfrac{Rg}{v_0^2},$$ $$2\theta=sin^{-1}(\dfrac{Rg}{v_0^2}),$$ $$2\theta=sin^{-1}(\dfrac{31\ m\cdot9.8\ \dfrac{m}{s^2}}{(18\ \dfrac{m}{s})^2})=70^{\circ},$$ $$\theta=\dfrac{70^{\circ}}{2}=35^{\circ}.$$