# Trajectory Motion

by zellster87
Tags: motion, trajectory
 P: 1,754 So we'll need to equations ... $$x=(v_0\cos\theta)t$$ Solve for initial velocity in the x, then plug that in $$y=(v_0\sin\theta)t-\frac{1}{2}gt^2$$ and solve for t.
 P: 1,754 Trajectory Motion Solving for velocity in the x ... $$v_0=\frac{x}{t\cos\theta}$$ Plugging into y ... $$y=\frac{x}{t\cos\theta}(\sin\theta)t-\frac{1}{2}gt^2$$ In your problem ... "at the same height above the ground as it was shot" so y=0. Solve for t.