A body is thrown with speed v_0 at an angle α to the…

A body is thrown with speed $v_0$ at an angle $\alpha$ to the horizontal. Find the angle its velocity makes with the horizontal at a horizontal distance $x$ from the launch point.

Explanation of Method

Direction of motion at a given horizontal distance.

Step 1: Trajectory

From $x = v_0\cos\alpha\, t$ and $y = v_0\sin\alpha\, t - \tfrac12 g t^2$, eliminate $t = \dfrac{x}{v_0\cos\alpha}$: $y(x) = x\tan\alpha - \frac{g\,x^2}{2v_0^2\cos^2\alpha}$.

Step 2: Direction = slope of the path

The velocity is tangent to the trajectory, so $\tan\theta = dy/dx$. (Equivalently $\tan\theta = v_y/v_x = \dfrac{v_0\sin\alpha - gt}{v_0\cos\alpha}$ with the same $t$.)

\[ \frac{dy}{dx} = \tan\alpha - \frac{g\,x}{v_0^2\cos^2\alpha}. \]

\[\theta = \arctan\left(\tan\alpha - \frac{g\,x}{v_0^2\cos^2\alpha}\right)\]

Checks

$x = 0$: $ \theta = \alpha $ (launch direction) $\checkmark$

At the top, $ x = \tfrac{v_0^2\sin\alpha\cos\alpha}{g} $: $ \tan\theta = \tan\alpha - \tan\alpha = 0 $ (moving horizontally) $\checkmark$

At the landing point $ x = R = \tfrac{2v_0^2\sin\alpha\cos\alpha}{g} $: $ \tan\theta = -\tan\alpha $, so $ \theta = -\alpha $ $\checkmark$

$\alpha$ $\theta$ $x$ $\tan\theta = \tan\alpha - \dfrac{gx}{v_0^2\cos^2\alpha}$

The direction of motion at horizontal distance $x$ is the slope of the trajectory there (dashed tangent). Before the top $\theta$ is positive (rising), after it negative (falling); at the landing point $\theta = -\alpha$.

Result

\[\boxed{\,\theta = \arctan\left(\tan\alpha - \frac{g\,x}{v_0^2\cos^2\alpha}\right)\,}\]

Remark

Note. If "the distance $x$" means the landing point on the same horizontal plane, the answer is simply that the body hits the ground at the angle $\alpha$ below the horizontal (the last check).