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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 $(0, 1, 0)$?

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  • $\begingroup$ 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! $\endgroup$
    – Newbie
    Commented Jan 12, 2022 at 4:11
  • $\begingroup$ @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 $(0.2261, -0.90443, 0.36177)$. $\endgroup$
    – DBolt
    Commented Jan 12, 2022 at 4:22
  • $\begingroup$ So the only force acting on the particle is $\vec F=-mg\vec u$? $\endgroup$
    – Newbie
    Commented Jan 12, 2022 at 4:25
  • $\begingroup$ Yes and in this instance, mass is negligible as the projectile is represented as a point with no radius. $\endgroup$
    – DBolt
    Commented Jan 12, 2022 at 4:32
  • $\begingroup$ 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 $\endgroup$
    – Newbie
    Commented Jan 12, 2022 at 4:38

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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 $(\Delta \vec r\cdot \vec u)\vec u$.

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