Deriving a formula that an old physicist showed me.

  • Thread starter amidda1217
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  • #1

Homework Statement


I was shown a formula by a physicist I met through a friend when I asked for help with a homework problem. I needed to find y final for a motion in 2 dimensions problem. He showed me this:

Y1 = X1 * tan(Theta) + (Ay * X1^2) /( 2 * V0^2 * cos^2(Theta))

Y1 = Y final
X1 = X Final
Ay = Acceleration of Y
V0 = Initial Velocity

I would love to know where this comes from and how to derive it. The X1tan(theta) makes sense to me: that would just = y of a triangle with base of x1. But the cos^2 and X1 ^2 confuses me. Did those come from an identity?

Have any of you seen this before?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Take the simple components

uyvsinθ and uxvcosθ (vertical & horizontal)

When the projectile hits the ground y=0

or y = uyt-1/2gt2 (u=initial and g = acceleration or your Ay)

you can solve for t here.


Then you know the entire horizontal range x=uxt, you know ux and t from above, so simplify.

If you're having trouble, post your work.
 
  • #3
So I was able to derive the second half of the problem from 1/2at^2. The part that I am actually stuck on is the first chunk. I don't understand how ViT derives to XfTan(Theta)

I got T = Xf/VCos(Theta)

But somehow Viy*T = Xf Tan(Theta). Not sure how to derive that.
 
Last edited:
  • #4
rock.freak667
Homework Helper
6,230
31
So I was able to derive the second half of the problem from 1/2at^2. The part that I am actually stuck on is the first chunk. I don't understand how ViT derives to XfTan(Theta)

I got T = Xf/VCos(Theta)

But somehow Viy*T = Xf Tan(Theta). Not sure how to derive that.

Now that I think it through, you are relating y with x.

we have y=uyt-½gt2

and we know that x=uxt or x=(vcosθ)t so t = x/vcosθ

and now y = (vsinθ)t-½gt2

sub t into it and you should get it.
 
  • #5
Thank you very much for your help. I figured it out. Couldn't have done it without help. :smile:
 

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