Projectile Motion: Deriving 2g^2

[Klaz]

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.

 

[gneill]

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.

 

[Doc Al]

Step by step. What's the first derivative of v^2?

 

[Klaz]

Step by step. What's the first derivative of v^2?

First derivative would be 2v.

 

[Klaz]

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]

First derivative would be 2v.

If Vy=v^2-gt then the first would be dv/dt = 2v-g.

 

[gneill]

First derivative would be 2v.

That's not complete. In that equation v is a function of time, so apply the chain rule.

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.

 

[gneill]

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.

 

[PeroK]

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?

 

[Doc Al]

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.)

 

[Klaz]

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.

 

[Doc Al]

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?

 

[Klaz]

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?

 

[Doc Al]

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.

Let u=(v-gt)^1/2

Where did you get this?

Perhaps you meant to follow gneill's advice:

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.

 

[Klaz]

I couldn't follow what you were trying to do.Where did you get this?

Perhaps you meant to follow gneill's advice:

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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.