Horizontally Launched Projectile - Angle of Velocity Just Before Hitting Ground

Given the initial height $h$, the initial distance the projectile is above the ground, and the horizontal distance $d$ between the projectile's initial and final positions (from launch position to when it hits the ground), how do we find the angle of the a horizontally launched projectile's velocity vector just before it hits the ground?

Since we are not given the initial velocity that the projectile is launched at, we can see that since it was launched horizontally, there is no vertical component of the initial velocity. So, we can see that since velocity $v = \frac{\Delta x}{t}$ and $d=\Delta x$, we have that $v_0 = \frac{d}{t}$. This initial horizontal velocity is constant for the projectile since no horizontal forces are acting on the projectile (we are assuming that air resistance is negligible), so we have that $v_x = v_0 = \frac{d}{t}$.

Now, we notice that we are asked to find the angle $\theta$ of the projectile's final velocity. Since we know that the final velocity has both a vertical component $v_y$ and a horizontal component $v_x$, we can see that $\tan{\theta} = \frac{v_y}{v_x}$ or $\theta = {\tan}^{-1}\left(\frac{v_y}{v_x}\right)$.

Since we want to relate $v_y$ and $v_x$, we note that the vertical and horizontal motions of the projectile are independent and only connected by time $t$. So, we can use acceleration $a = \frac{v}{t}$ for the vertical motion of the projectile where $a = g$, the acceleration due to gravity, to get $g = \frac{v_y}{t}$. Solving for $t$, we find $t = \frac{v_y}{g}$.

To express $v_y$ in terms of $d$, $h$, and constant $g$, we can use the equation $v^2 = v_0^2 + 2a\Delta x$ that holds under constant acceleration such as in projectile motion. We get $v_y^2 = 0 + 2gh$, which gives us $v_y = \sqrt{2gh}$.

So, substituting $v_y = \sqrt{2gh}$ into $t = \frac{v_y}{g}$ gives us $t = \frac{\sqrt{2gh}}{g} = \sqrt{\frac{2h}{g}}$.

Relating this to $v_x = \frac{d}{t}$, we find $v_x = \frac{d}{\sqrt{\frac{2h}{g}}} = d\sqrt{\frac{g}{2h}}$.

Thus, we have $\theta = {\tan}^{-1}\left(\frac{v_y}{v_x}\right) = {\tan}^{-1}\left(\frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}\right) = {\tan}^{-1}\left(\frac{2h}{d}\right)$ as our answer.

The interesting thing to note is that we actually don't need to know the initial horizontal velocity of the projectile since the distance it travels can help us find the initial velocity.

An extension to the problem is to consider if the projectile was not launched horizontally; that is, the initial velocity of the projectile had both a vertical and a horizontal component. We would still be able to find $v_{x_0}$, but we would need to know the vertical component of the initial velocity $v_{y_0}$ to solve the problem.

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