Parabola of a projectile?

In summary, the conversation is about someone seeking help on an equation to show the parabola of a projectile. They have determined that lateral and vertical velocities are needed, but are stuck after that. The person refuses to look up the solution and has only come up with an expression involving delta x and delta y. They need to factor in gravity and deceleration and are new to physics. The other person suggests starting with the equations x = utcosθ and y = utsinθ - 1/2gt^2 and provides a resource for further understanding of projectile motion.
  • #1

CallMeDirac

46
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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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  • #2
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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  • #3
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
 
  • #4
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
 
  • #5
CallMeDirac said:
That didnt work
You're short of a few dollars! You deleted the dollar signs that delimit the Latex.
 
  • #6
PeroK said:
You're short of a few dollars! You deleted the dollar signs that delimit the Latex.
$$ \Delta x^2 + \Delta y + h $$
?
 
  • #7
CallMeDirac said:
$$ \Delta x^2 + \Delta y + h $$
?
yes
 
  • #8
CallMeDirac said:
$$ \Delta x^2 + \Delta y + h $$
?
That's not an expression I can make any sense of.
 
  • #9
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:
 

What is a "parabola of a projectile"?

The parabola of a projectile refers to the curved path that a projectile follows when launched into the air. It is a result of the combination of the projectile's initial velocity and the force of gravity pulling it towards the ground.

What factors affect the shape of the parabola?

The shape of the parabola is affected by the initial velocity, launch angle, and the acceleration due to gravity. These factors determine the height, width, and symmetry of the parabola.

How is the parabola of a projectile calculated?

The parabola of a projectile can be calculated using the equations of motion, which take into account the initial velocity, launch angle, and acceleration due to gravity. Alternatively, it can also be graphically represented by plotting the position of the projectile at different points in time.

What is the significance of the parabola of a projectile?

The parabola of a projectile is significant because it helps us understand the motion of objects in projectile motion. It also has practical applications in fields such as physics, engineering, and sports, where the trajectory of a projectile needs to be accurately predicted.

Can the parabola of a projectile be affected by external factors?

Yes, the parabola of a projectile can be affected by external factors such as air resistance and wind. These factors can alter the shape and trajectory of the parabola, making it more complex to calculate and predict.

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