How can I determine the parabola of a projectile?

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    Parabola Projectile
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SUMMARY

The discussion focuses on determining the parabolic trajectory of a projectile using the equations of motion. Key equations mentioned include $$x = ut\cos \theta$$ and $$y = ut\sin \theta - \frac{1}{2}gt^2$$, where 'u' represents initial velocity, 't' is time, and 'g' is the acceleration due to gravity. Participants emphasize the need to account for gravity and deceleration in the calculations, and the importance of using LaTeX for formatting equations. The conversation highlights the necessity of understanding both lateral and vertical velocities to accurately model projectile motion.

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  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with the equations of motion in two dimensions.
  • Knowledge of LaTeX for formatting mathematical expressions.
  • Basic understanding of the effects of gravity on motion.
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  • Study the derivation of the equations of motion for projectile trajectories.
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  • Practice using LaTeX to format and present mathematical equations clearly.
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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
 
  • #10
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:$$x = ut\cos \theta, \ y = ut\sin \theta -\frac{1}{2}gt^2$$
 
  • #11
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
 
  • #12
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:
 

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