Galilean relativity in projectile motion

Question

Consider a reference frame $S^'$ moving in the initial direction of motion of a projectile launched at time, $t=0$. In the frame $S$ the projectile motion is:

$$x=u(cos\theta)t$$

$$y=u(sin\theta)t-\frac{g}{2}t^2$$

I know that that at $y_{max}$, $\frac{dy}{dt}=0$ so using this I find that $$t_{y_{max}}=\frac{usin\theta}{g}$$

so therefore: $$y_{max}=\frac{2u^2sin\theta}{g}$$

I know that when the particle lands at $y_{bottom}=0$ the distance in the $x$ direction is $$x_{y_{bottom}}=\frac{2u^2sin^2\theta}{g}$$

but I am confused about how to describe the motion for the particle in $S'$ frame.

Answer 1

In the $S'$ frame, your variables are $x' = x - t\cdot u \cos\theta $ and $y' = y - t\cdot u \sin\theta$. If you do the change of variable, you get that the motion now is described by

$$x' = 0$$ $$y' = -\frac{g}{2}t^2$$

So in your new frame of reference you have vertical free fall from rest.

This is not very helpful in finding out when or where does the projectile hits the ground, but is very relevant if you want to know where will the projectile be after releasing it from a plane moving at constant velocity: right below it all the time. Disregarding air resistance, of course.

EDIT The system with a prime is moving with velocity $(u \cos\theta, u\sin\theta)$, so if you have a velocity in the unprimed system, to convert it to the primed system, you have to substract the velocity of the origin:

$$\vec{v'} = \vec{v} - (u \cos\theta, u\sin\theta)$$

Integrating this, you can get the relation for the position vector:

$$\vec{r'} = \vec{r} - (u \cos\theta, u\sin\theta)t + \vec{r}_0$$

where $\vec{r}_0$ is the position of the origin of the primed system for $t=0$. Both systems share origin for $t=0$, so $\vec{r}_0=\vec{0}$.

Now replace $\vec{r'}=(x',y')$ and $\vec{r}=(x,y)$ and you will get the equations above.