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Kinematics - trajectory formula

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1. Homework Statement
given :
1409395.jpg
i need to find the trajectory formula

2. Homework Equations
i'm not sure if to use :
6687766.jpg


3. The Attempt at a Solution
6325967.jpg

I tried different options with the trigonometric identities that I have written before:
8541126.jpg

thanks
 

PeroK

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3. The Attempt at a Solution
6325967.jpg
If ##x## and ##y## are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.
 
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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?

9133966.jpg
 

PeroK

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do you mean?

9133966.jpg
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.
 
7
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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

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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}##.
 
7
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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

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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!
 

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