Can someone please help me to eliminate $ t$ from functions

Question

I have 2 functions and I don't know how to eliminate variable $t$ from them. Could someone please help me with this...

$x(t)=Vt-\frac{bt^2 \sin(\omega)}{2}$ $(1)$

$y(t)=\frac{bt^2\cos(\omega)}{2}$ $(2)$

Is it possible to get $y(x)$ without variable $t$ in it...

My work :

when I substitute $bt^2=\frac{Vt-x}{sin(\omega)}$ from $(1)$

I get $y(x)=\frac{(vt-x)cos(\omega)}{sin(\omega)}$

Is it possible to get $y(x)$ without the variable $t$ in it ? If so could you please help me with it, Thank you in advance. :)

Answer 1

We can obtain $t$ from

$$y(t)=\frac{bt^2\cos(\omega)}{2}\implies t=\sqrt{\frac{2y}{b \cos \omega}}$$

and substitute in $x(t)=Vt-\frac{bt^2 \sin(\omega)}{2}$.

Or, as an alternative, obtain $t$ from

$$x(t)=Vt-\frac{bt^2 \sin(\omega)}{2}\implies \frac{b \sin(\omega)}{2}t^2-Vt+x=0$$

and substitute in $y(t)=\frac{bt^2\cos(\omega)}{2}$.

Answer 2

If $b$ is accleration due to gravity, then

$$ \ddot y(t) = -b$$

$$ \dot x(t)=V \cos \omega \tag1 $$

$$ x(t)=V t \cos \omega \, \tag2 $$

$$ \dot y(t)= -b t + V\sin \omega \tag3$$

$$ y(t)= -b t^2/2 + Vt \sin \omega \tag4 $$

where I assumed the projectile starts at the origin

Now solve for $t$ from 2) and plug into (4)

Answer 3

If you try:

$x(t)=Vt-\frac{bt^2 \sin(\omega)x}{2}$

$t = \frac {V \pm \sqrt{V^2 -2b\sin(\omega)x}}2$

$y=\frac{bt^2\cos(\omega)}{2}$

$=\frac {b(\frac {V^2 - 2b\sin(\omega)x}4 \pm V\sqrt{V^2 -2b\sin(\omega)x}+bV^2)\cos(\omega)}2$

Ugly, but doable. Probably has a simpler form.