Kinematics - trajectory formula

1. Nov 7, 2016

quas

1. The problem statement, all variables and given/known data
given : i need to find the trajectory formula

2. Relevant equations
i'm not sure if to use :

3. The attempt at a solution

I tried different options with the trigonometric identities that I have written before:

thanks

2. Nov 7, 2016

PeroK

If $x$ and $y$ are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.

3. Nov 7, 2016

do you mean?

4. Nov 7, 2016

PeroK

No, I meant:

$\frac{y}{a} = \cos(2 \omega t) =$ something to do with $\sin(\omega t) =$ something to do with $\frac{x}{a}$

You're terribly over-complicating this.

5. Nov 7, 2016

quas

Ok I can write $\frac{y}{a} =1-2sin^2(\omega t)$, also to $\frac{x^2}{a^2}\ = sin^2(\omega t)$ and then insert $sin^2(\omega t)$
to the equation $\frac{y}{a} =1-2sin^2(\omega t)$ . would it be correct?

6. Nov 7, 2016

PeroK

You seem to have a mental block about replacing $\sin(\omega t)$ by $\frac{x}{a}$.

7. Nov 7, 2016

quas

first of all thanks for your help it's not taken for granted
Last try:
in the end I will get : $\frac{2x^2}{a^2}+ \frac{y}{a}=1$ then : $y= - \frac{2}{a}x^2+a$ and thats parabola figure . right ?

8. Nov 7, 2016

It is!