Special relativity in projectile motion I'm trying to describe projectile motion wrt a reference frame $S'$ that moves with velocity $v=0.6c$ in the $x$ direction wrt to the reference frame $S$ where the projectile was launched. I know projectile motion equations and Lorentz transformations but I'm kind of struggling to understand what is it that $S'$ sees.
In $S$ I know $$x(t) = x_0 + v_o\cos(\alpha)t$$ $$y(t) = y_0 + v_o\sin(\alpha)t-\frac{g}{2}t^2$$
And Lorentz transformations are
$$ 
\left[ {\begin{array}{c}
    x' \\
    y' \\
    t'
  \end{array} } \right]
= \underbrace{\left[ {\begin{array}{ccc}
    \gamma & 0 &-v\gamma\\
     0 & 1 & 0 \\
    -v\gamma/c^2 & 0 & \gamma \\
  \end{array} } \right]}_{L_{xv}}
\left[ {\begin{array}{c}
    x(t) \\
    y(t) \\
    t
  \end{array} } \right]$$
Now here are my questions,

*

*Is the last equation correct? (Correct in the sense that I simply have to substitute the expressions for $x(t)$ and $y(t)$) If yes, then $x',y'$ would depend on $t$ and shouldn't they depend on $t'$ instead?

*I am quite intrigued by the angle $\alpha$. Shouldn't I also apply some sort of transformation $\alpha\rightarrow\alpha'$? because $S'$ won't see the same angle that $S$ does, right?

*Also should I make a velocity transformation for the initial velocity and somehow account for it in $S'$? If that's the case how would I need to approach and add this to the equations of $S'$?

*Lastly, if I were to calculate the range and maximum height that $S'$ sees (which in $S$ are $R = x_0+\frac{v_0^2}{g}\sin(2\alpha)$ and $h_{max} = y_0 + v_0^2\frac{\sin(\alpha)^2}{2g}$ respectively) since $y' = y$ then $h_{max}':$ maximum height seen from $S'$, should be the same right? And for $R':$ range seen from $S'$, would I need to find $t'$ (or $t$ not even sure) such that $y' = 0$ and substitute this value of $t'$ (or $t$) in $x'$ to get $R'$? Or would it be possible I use some transformation for $\alpha$, $v_0$ and $x_0$ and get $R'$ that way?

Any information or help would be greatly appreciated.
 A: Yes, your matrix equation is correct, although we normally put the time component at the top, not the bottom.
You don't actually need to worry about transforming $g, v_o, \alpha$. Once you have equations for $x', t'$ in terms of $x, t$, you can substitute in
$$x = x_0 + v_o\cos(\alpha)t$$
and then eliminate $t$, which will let you write $x'$ as a function of $t'$.
For small $v_o$, the projectile's motion in the $S'$ frame is almost a straight line going backwards at 0.6c. If $v_o$ is large, then your basic projectile motion equations for $x, y$ in $S$ are probably inadequate, unless you happen to have a gigantic plane with uniform gravity. ;)

I won't do the substitution I mentioned above, but here's a simpler example to give you the general idea. Instead of parabolic motion, the projectile has simple uniform motion:
$$x = ut$$
with $y=0$, so we can ignore $y$ and $y'$.
The Lorentz transformation gives us
$$x' = \gamma(x-vt)$$
$$ct' = \gamma(ct-vx/c)$$
We have $v=\frac35 c$, therefore $\gamma=\frac54$.
So
$$x'=\frac54 x - \frac34 ct$$
$$ct'=\frac54 ct - \frac34 x$$
Substituting $x = ut$,
$$x'=\left(\frac54 u - \frac34 c\right)t$$
$$ct'=\left(\frac54 c - \frac34 u\right)t$$
Thus
$$t = \frac{ct'}{\frac54 c - \frac34 u}$$
Substituting that into our last equation for $x'$ we get
$$x'=\left(\frac{\frac54 u - \frac34 c}{\frac54 c - \frac34 u}\right)ct'$$
or
$$x'=\left(\frac{5u - 3c}{5c - 3u}\right)ct'$$
We can write that as
$$x'=u't'$$
where
$$u'=\left(\frac{5u - 3c}{5c - 3u}\right)c$$
Note that for small $u$, $u'\approx u-v$

The Lorentz transformation is symmetrical, and it can often be useful to work with the inverse form,
$$x = \gamma(x'+vt')$$
$$ct = \gamma(ct'+vx'/c)$$
Here's a more general way to find $u'$.
$$x' = \gamma(x-vt)$$
$$t' = \gamma(t-vx/c^2)$$
Now $u'=x'/t'$, thus
$$u'=\frac{x-vt}{t-vx/c^2}$$
Substituting $x=ut$
$$u'=\frac{ut-vt}{t-uvt/c^2}$$
$$u'=\frac{u-v}{1-uv/c^2}$$
which can also be written in this form:
$$\frac{u'}c=\frac{\frac uc-\frac vc}{1-\frac uc \frac vc}$$
And due to symmetry,
$$u=\frac{u'+v}{1+uv/c^2}$$
This is the well-known law of addition of relativistic velocities.
