Projectile Motion Equation Need Help Re-arranging Please!

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Hey, this is my first time posting here, I just found this site in a Google search.

I've been doing a projectile motion question, and I know what I am doing and solving for and such, but I am having great difficulties re-arranging the equation I need. The equation looks like so:
[tex]\Delta[/tex]d[tex]_{}y[/tex] = v[tex]_{2y}[/tex][tex]\Delta[/tex]t + [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]

The variable I need to solve for is time ([tex]\Delta[/tex]t), but I'm having vast difficulties in doing so. I know I've solved this before, but am not entirely sure how I did it. If this makes no sense, I will post the whole question so I can be told what I'm doing wrong.

Thanks in advance for the help, I would really appreciate it!

EDIT: I think I'm doing this posting thing all wrong...sorry for my noob-ness.

Ninjaedit: the speed shouldn't be squared, it should be v2 (as in second) y...subscript doesn't work for me properly apparently

EDIT AGAIN: Okay, so in reading around a little bit more, I noticed I should list what I went through to try to rearrange this equation.
1. [tex]\Delta[/tex]d[tex]_{}y[/tex] - v[tex]_{2y}[/tex][tex]\Delta[/tex]t = [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
2. [tex]\frac{[tex]\Delta[/tex]d[tex]_{}y[/tex] - v[tex]_{2y}[/tex][tex]\Delta[/tex]t}{0.5g}[/tex] = [tex]\Delta[/tex]t[tex]^{2}[/tex]

And that's where I got stumped. How would I isolate this variable? Is it do-able?
 
Last edited:

Answers and Replies

  • #2
Nabeshin
Science Advisor
2,205
16
Hey, this is my first time posting here, I just found this site in a Google search.

I've been doing a projectile motion question, and I know what I am doing and solving for and such, but I am having great difficulties re-arranging the equation I need. The equation looks like so:
[tex]\Delta[/tex]d[tex]_{}y[/tex] = v[tex]_{2y}[/tex][tex]\Delta[/tex]t + [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]

The variable I need to solve for is time ([tex]\Delta[/tex]t), but I'm having vast difficulties in doing so. I know I've solved this before, but am not entirely sure how I did it. If this makes no sense, I will post the whole question so I can be told what I'm doing wrong.

Thanks in advance for the help, I would really appreciate it!

EDIT: I think I'm doing this posting thing all wrong...sorry for my noob-ness.

Ninjaedit: the speed shouldn't be squared, it should be v2 (as in second) y...subscript doesn't work for me properly apparently
'Ello, and welcome to PF!

In general, the solution to this is a quadratic which, depending on your math background, can look really sticky or really neat. If you re-write the equation like this:
[tex]\frac{1}{2}g\Delta t^{2}+v_{2y}\Delta t - \Delta d_{y}=0[/tex]
You can see it's in the form of a quadratic...
[tex]at^{2}+bt+c=0[/tex]

And I'll assume you're familiar with the quadratic formula...

Edit:

There is a special case of this when [tex]\Delta d_{y}=0[/tex] which isn't a quadratic (well it is, but one solution is always t=0).
 
  • #3
8
0
'Ello, and welcome to PF!

In general, the solution to this is a quadratic which, depending on your math background, can look really sticky or really neat. If you re-write the equation like this:
[tex]\frac{1}{2}g\Delta t^{2}+v_{2y}\Delta t - \Delta d_{y}=0[/tex]
You can see it's in the form of a quadratic...
[tex]at^{2}+bt+c=0[/tex]

And I'll assume you're familiar with the quadratic formula...

Edit:

There is a special case of this when [tex]\Delta d_{y}=0[/tex] which isn't a quadratic (well it is, but one solution is always t=0).
Aaah, thank you very much for the reply! And yes, I am familiar with the Quadratic Formula, but haven't used it in a year. I made the mistake of stopping math at Grade 11 U level, in the same semester as my Grade 11 U Physics, and then taking Grade 12 U Physics, after having no math for a year.

tl;dr what's the Quadratic equation again?

Thanks again!

EDIT: Nvm, got it, thanks!
 

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