# Kinematics - trajectory formula

#### quas

1. Homework Statement
given :
i need to find the trajectory formula

2. Homework 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

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#### PeroK

Homework Helper
Gold Member
2018 Award
3. The Attempt at a Solution
If $x$ and $y$ are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.

#### quas

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

#### PeroK

Homework Helper
Gold Member
2018 Award
do you mean?

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.

#### quas

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.
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?

#### PeroK

Homework Helper
Gold Member
2018 Award
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?
You seem to have a mental block about replacing $\sin(\omega t)$ by $\frac{x}{a}$.

#### quas

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

#### PeroK

Homework Helper
Gold Member
2018 Award
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 ?
It is!

"Kinematics - trajectory formula"

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