How can I determine the parabola of a projectile?

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CallMeDirac
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I have been puzzling over an equation that could be made to show the parabola of a projectile.
So far I have determined that the lateral and vertical velocities are needed, the lateral velocity should determine the x² function but after that I am stuck.

To specify I refused to look this up as that would be admitting defeat so I figured this was second best.
 
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CallMeDirac said:
Summary:: I need help on this

I have been puzzling over an equation that could be made to show the parabola of a projectile.
So far I have determined that the lateral and vertical velocities are needed, the lateral velocity should determine the x² function but after that I am stuck.

To specify I refused to look this up as that would be admitting defeat so I figured this was second best.
You need to post what you've done so far. Try to use Latex if you can:

https://www.physicsforums.com/help/latexhelp/

For example, I guess you have: $$x = ut\cos \theta, \ y = ut\sin \theta -\frac{1}{2}gt^2$$

If you reply to this post you should see what I typed to get those formatted equations.
 
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PeroK said:
You need to post what you've done so far. Try to use Latex if you can:

https://www.physicsforums.com/help/latexhelp/

For example, I guess you have: $$x = ut\cos \theta, \ y = ut\sin \theta -\frac{1}{2}gt^2$$

If you reply to this post you should see what I typed to get those formatted equations.
\delta x^2 + \delta y + h

Is all I have but I need to account for gravity and find the rate of deceleration
 
CallMeDirac said:
\delta x^2 + \delta y + h

Is all I have but I need to account for gravity and find the rate of deceleration
That didnt work
 
CallMeDirac said:
That didnt work
You're short of a few dollars! You deleted the dollar signs that delimit the Latex.
 
PeroK said:
You're short of a few dollars! You deleted the dollar signs that delimit the Latex.
$$ \Delta x^2 + \Delta y + h $$
?
 
CallMeDirac said:
$$ \Delta x^2 + \Delta y + h $$
?
yes
 
CallMeDirac said:
$$ \Delta x^2 + \Delta y + h $$
?
That's not an expression I can make any sense of.
 
PeroK said:
You're short of a few dollars! You deleted the dollar signs that delimit the Latex.

So far I have

$$ \Delta x^2 + \Delta y + h $$

$$ \Delta x^2 $$ being the denominator in the slope and $$ \Delta y $$ being the numerator ( rise/ run for slope) and H being the height from which it is fired, but I need to factor in gravity and deceleration
 
CallMeDirac said:
So far I have

$$ \Delta x^2 + \Delta y + h $$

$$ \Delta x^2 $$ being the denominator in the slope and $$ \Delta y $$ being the numerator ( rise/ run for slope) and H being the height from which it is fired, but I need to factor in gravity and deceleration
That makes no sense.

If you are trying to do what I think you are doing, you need to start with this:$$x = ut\cos \theta, \ y = ut\sin \theta -\frac{1}{2}gt^2$$
 
PeroK said:
If you are trying to do what I think you are doing, you need to start with this:$$x = ut\cos \theta, \ y = ut\sin \theta -\frac{1}{2}gt^2$$

Can you explain each part.
Sorry, I am a bit new to physics
 
CallMeDirac said:
So far I have

$$ \Delta x^2 + \Delta y + h $$

$$ \Delta x^2 $$ being the denominator in the slope and $$ \Delta y $$ being the numerator ( rise/ run for slope) and H being the height from which it is fired, but I need to factor in gravity and deceleration
The real Paul Dirac would never have done anything like that! :wink: