# Displacement of a Projectile Along a Vector

I've been stuck on this problem for more time than I'd like to admit, and I feel like I'm overcomplicating the solution, or just not understanding the concept.

I have a projectile in motion in three-dimensional space that starts at some initial point $$\vec{r_i}$$ and impacts a surface at point $$\vec{r_f}$$. I have the initial velocity on all axes as $${v_i}$$, I have the gravity as scalar $${g}$$ and I also have the vector opposite the direction of gravity, which I'll call $$\vec{u}$$. Multiplying $${g}$$ and $$\vec{u}$$ gives me the gravitational acceleration vector $$\vec{a}$$.

Typically, it's assumed that $$\vec{u} = (0, 1, 0)$$, so vertical displacement could be acquired by $$\Delta{y} = {r_f}_y - {r_i}_y$$ or as $$\Delta{y} = \frac{1}{2}{g}{t^2}$$ if you have time, which I don't.

In some instances during this problem, $$\vec{u}$$ is represented as a normalized vector that is not equivalent to the y-axis vector $$(0, 1, 0)$$.

Given that, how can I find vertical displacement in along $$\vec{u}$$ instead of the assumed y-axis vector $$(0, 1, 0)$$?

• By vertical displacement you mean in the direction of $(x,y,z)=(0,1,0)$? Also, I recommend $\vec r$ instead of $x$ to denote location (position) since you're discussing the $x$ and $y$ coordinates of the Cartesian system! Commented Jan 12, 2022 at 4:11
• @Newbie Updated those variables, thank you! I mean displacement along the vector $\vec{u}$, representing the vector opposite the direction of gravity, which could be $(0, 1, 0)$ but it could also be some other normalized vector such as $(0.2261, -0.90443, 0.36177)$. Commented Jan 12, 2022 at 4:22
• So the only force acting on the particle is $\vec F=-mg\vec u$? Commented Jan 12, 2022 at 4:25
• Yes and in this instance, mass is negligible as the projectile is represented as a point with no radius. Commented Jan 12, 2022 at 4:32
• The total displacement vector is $\Delta \vec r=\vec r_{\rm f}-\vec r_{\rm i}$. Assuming $\vec u$ is normalized to 1, the magnitude of displacement along $\vec u$ is $\Delta \vec r\cdot \vec u$ and the displacement vector along $\vec u$ is $(\Delta \vec r\cdot \vec u)\vec u$. Tell me what I'm missing because I feel you already have the start and end point of the trajectory and don't need time Commented Jan 12, 2022 at 4:38

The total displacement vector is $$\Delta \vec r=\vec r_{\rm f}-\vec r_{\rm i}$$. Assuming $$\vec u$$ is normalized to 1, the magnitude of displacement along $$\vec u$$ is $$\Delta \vec r\cdot \vec u$$ and the displacement vector along $$\vec u$$ is $$(\Delta \vec r\cdot \vec u)\vec u$$.