Projectile Motion: Deriving 2g^2

AI Thread Summary
The discussion revolves around deriving the equation d²(v²)/dt² = 2g² for projectile motion. Participants emphasize the importance of correctly applying the chain rule when differentiating v² with respect to time. The correct approach involves starting with the vertical velocity equation, v = v₀ - gt, and then squaring it to find v². After differentiating v² twice, the result confirms the desired expression of 2g². The conversation highlights common misunderstandings in applying calculus to physics problems.
Klaz
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Homework Statement


Show that for a projectile d^2 (v^2)/dt^2 = 2g^2

Homework Equations

The Attempt at a Solution


Since we are only dealing with the y component for this. 2nd derivative would be 2. But since gravity acts on it then d^2 (v^2)/dt^2 = 2g. I don't get how to get to 2g^2. Can someone help me please.
 
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Klaz said:

Homework Statement


Show that for a projectile d^2 (v^2)/dt^2 = 2g^2

Homework Equations

The Attempt at a Solution


Since we are only dealing with the y component for this. 2nd derivative would be 2. But since gravity acts on it then d^2 (v^2)/dt^2 = 2g. I don't get how to get to 2g^2. Can someone help me please.
Start with a general expression for the vertical velocity of a projectile.
 
Step by step. What's the first derivative of v^2?
 
Doc Al said:
Step by step. What's the first derivative of v^2?
First derivative would be 2v.
 
gneill said:
Start with a general expression for the vertical velocity of a projectile.
Vertical velocity = v - at. I am still confuse on what to do with this equation. Since dv/dt = a.
 
Klaz said:
First derivative would be 2v.
If Vy=v^2-gt then the first would be dv/dt = 2v-g.
 
Klaz said:
First derivative would be 2v.
That's not complete. In that equation v is a function of time, so apply the chain rule.

Klaz said:
Vertical velocity = v - at. I am still confuse on what to do with this equation. Since dv/dt = a.
that would be ##v = v_o - g t##. Square both sides so that you have an expression for v2. Then do your derivatives. Remember that dv/dt = g.
 
Klaz said:
If Vy=v^2-gt then the first would be dv/dt = 2v-g.
I don't know where you got that expression for Vy. It's not correct.
 
Klaz said:
Vertical velocity = v - at. I am still confuse on what to do with this equation. Since dv/dt = a.

It's your maths that's letting you down here. I don't think you understand what is being asked. You are asked to differentiate ##v^2##. Not ##v##.

I would first work out what ##v^2## is. That seems logical to me: you are asked to differentiate ##v^2##, with respect to time (twice). So, let's first have ##v^2## as a function of ##t##.

##v^2 = \dots ##

Can you do that?
 
  • #10
Klaz said:
First derivative would be 2v.
As gneill already pointed out, that is not complete.

What you found is the first derivative with respect to v. But what you need is the first derivative with respect to t. So, keep going. (Think chain rule.)
 
  • #11
If I apply chain rule. Let u=(v-gt)^1/2. Then : dv/dt = (u)^1/2
dv/dt = du/dt × dv/du = -g/2 (v-gt)^1/2. Am I on the right track? Then my next step would be to differentiate my answer again.
 
  • #12
Klaz said:
If I apply chain rule. Let u=(v-gt)^1/2. Then : dv/dt = (u)^1/2
dv/dt = du/dt × dv/du = -g/2 (v-gt)^1/2. Am I on the right track? Then my next step would be to differentiate my answer again.
Why not start with what you were given?

Let u = v^2
du/dt = du/dv * dv/dt.

Hint: What is dv/dt for a projectile?
 
  • #13
Doc Al said:
Why not start with what you were given?

Let u = v^2
du/dt = du/dv * dv/dt.

Hint: What is dv/dt for a projectile?
du/dt = 2v * -g = -2vg.
2nd derivative:
du/dt = -2g * - g = 2g^2.
Thank you. Can I ask one more question. What is wrong with my first attempt using the chain rule? Can i still get the same answer if I kept going with my solution?
 
  • #14
Klaz said:
What is wrong with my first attempt using the chain rule? Can i still get the same answer if I kept going with my solution?
I couldn't follow what you were trying to do.

Klaz said:
Let u=(v-gt)^1/2
Where did you get this?

Perhaps you meant to follow gneill's advice:
gneill said:
that would be ##v = v_o - g t##. Square both sides so that you have an expression for ##v^2##.

You can try that again and you should get the same answer.
 
  • #15
Doc Al said:
I couldn't follow what you were trying to do.Where did you get this?

Perhaps you meant to follow gneill's advice:

You can try that again and you should get the same answer.
Yes I followed his advice. I'll give it one more try using that. Thank you.
 
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