How to Derive Time Taken in Projectile Motion?

[jisbon]

Homework Statement

When 2 projectile motions are launched with velocity v, one at angle x and the other one at angle 90-x Derive an expression for the time difference in terms of range, x and g

Relevant Equations

R = v^2 sin2x / g

The below attempt will look messy, but I'm really kind of stuck as how to proceed on here. Thanks for any advice.

 

[BvU]

You are making things difficult for yourself. $S_{1,y} = S_{2, y}$ is correct, but since they are both equal to zero, one factor of $t_i$ drops out...

 

[jisbon]

You are making things difficult for yourself. $S_{1,y} = S_{2, y}$ is correct, but since they are both equal to zero, one factor of $t_i$ drops out...

Do I drop out t1 from the beginning where s1y=s2y, or do I drop it out at my last step?

 

[BvU]

You have $$S_{1,y} = 0 \quad \Rightarrow v \sin\theta_1\; t_1 + {\scriptstyle {1\over 2}}\,g\,t_1^2 = 0 \Rightarrow t_1 = ... $$I would drop the factor $t_1$ straightaway (it is the trivial one of two solutions for S=0).

 

[jisbon]

You have $$S_{1,y} = 0 \quad \Rightarrow v \sin\theta_1\; t_1 + {\scriptstyle {1\over 2}}\,g\,t_1^2 = 0 \Rightarrow t_1 = ... $$I would drop the factor $t_1$ straightaway (it is the trivial one of two solutions for S=0).

Alright it looks cleaner now, but now I have to get rid of v. If I used vcos theta t1 = R , this means v = R/cos theta t1, which I suspect it will be problematic. Any ideas on proceeding on from here?

 

[BvU]

You have your relevant equation for R

And $\sin(90^\circ-\theta)$ looks really ugly to me . Alternative ?

So does $\sin (2\theta)$

 

[jisbon]

You have your relevant equation for R

And $\sin(90^\circ-\theta)$ looks really ugly to me . Alternative ?

So does $\sin (2\theta)$

Hi, as per your hints:

I think I might have messed up as the final answer provided is drastically different from this. Any ideas on where I might have messed something up?

 

[BvU]

My route was to write v as a square root in the expression for $t_1-t_2$

 

[jisbon]

OK will do that and get back to you

 

[jisbon]

My route was to write v as a square root in the expression for $t_1-t_2$

Did you mean something like this?

 

[haruspex]

Did you mean something like this?

You seem to have gone wrong in the second line, using something like t=2v_x (?)
I preferred the top half of your post #7, down to where you had cos²+...+sin². With that hint, try to simplify differently.

 

[jisbon]

You seem to have gone wrong in the second line, using something like t=2v_x (?)
I preferred the top half of your post #7, down to where you had cos²+...+sin². With that hint, try to simplify differently.

In the cos^2 theta + sin^2 theta equation though, I have already subbed v in as the square root of... already though. The only thing I can proceed on seems to be only simplifying it which I have done

 

[BvU]

as the final answer provided

Is information we don't have. Do we keep guessing or can you share it ?

Something with $\left ( {\scriptstyle \sqrt{\tan\theta} }- {1\over \sqrt{\tan\theta}} \right )$ perhaps ?

 

[jisbon]

Yep with square root r/2g in front

 

[BvU]

Following up post #8:
$$t_1-t_2 = {2v\over g} \left (\cos\theta-\sin\theta\right ) \\ R = {2v^2 \sin\theta\cos\theta\over g }\Rightarrow v = \sqrt{Rg\over 2\sin\theta\cos\theta\ } $$
substitute and I get a $\sqrt {2R/g}$ in front, not $\sqrt {R/2g}$ ? [edited]

 

[haruspex]

The only thing I can proceed on seems to be only simplifying it which I have done

You simplified it in a particular way. I was suggesting a different simplification.

But now you have disclosed the desired form of the answer, it is clear that you would do better not to square t₁-t₂ in the first place. Just sub in the expression for v and simplify terms like $\frac{\cos(\theta)}{\sqrt{\sin(\theta)\cos(\theta}}$ (as @BvU was suggesting in post #8).

 

[jisbon]

You simplified it in a particular way. I was suggesting a different simplification.

But now you have disclosed the desired form of the answer, it is clear that you would do better not to square t₁-t₂ in the first place. Just sub in the expression for v and simplify terms like $\frac{\cos(\theta)}{\sqrt{\sin(\theta)\cos(\theta}}$ (as @BvU was suggesting in post #8).

Following up post #8:
$$t_1-t_2 = {2v\over g} \left (\cos\theta-\sin\theta\right ) \\ R = {2v^2 \sin\theta\cos\theta\over g }\Rightarrow v = \sqrt{Rg\over 2\sin\theta\cos\theta\ } $$
substitute and I get a $\sqrt {2R/g}$ in front, not $\sqrt {R/2g}$ ? [edited]

As per both of your instructions, I have done this:

Now I know this is more of a Maths question now rather than a physics question, but how do I get rid of the 2sin theta cos theta in this case? Thanks

 

[haruspex]

As per both of your instructions, I have done this:

View attachment 248293
Now I know this is more of a Maths question now rather than a physics question, but how do I get rid of the 2sin theta cos theta in this case? Thanks

Can you see how to get from that the term I wrote in post 16, and a similar one with sin instead of cos in the numerator?
What happens in each of those if you bring everything inside the square root?

 

[jisbon]

Can you see how to get from that the term I wrote in post 16, and a similar one with sin instead of cos in the numerator?
What happens in each of those if you bring everything inside the square root?

I'm not sure if its wrong, but I got the following:

From my previous post:

$t1 - t2$ = $2\sqrt{\frac{R}{2sin\theta cos\theta}}\div\sqrt{g}\times(cos\theta-sin\theta)$

If I combine everything in square root, will this be the answer?

= $2\sqrt{\frac{R(cos^2\theta-2sin\theta cos\theta+sin^2\theta)}{2sin\theta cos\theta g}}$

 

[BvU]

You have a tendency to make things complicated.

What is $\cos\theta/\sqrt{\cos\theta\sin\theta}$ ?

What is $\sin\theta/\sqrt{\cos\theta\sin\theta}$ ?

Oh, and a tip: use \sin and \cos and \tan in $\LaTeX$

 

[haruspex]

What happens in each of those if you bring everything inside the square root?

If I combine everything in square root, will this be the answer?

I tried to word that carefully ("each"). Let me try again...
Just consider one term, $\frac{\cos(\theta)}{\sqrt{\cos\theta\sin\theta}}$, say. What happens to that if you bring everything inside the square root.

 

[jisbon]

I tried to word that carefully ("each"). Let me try again...
Just consider one term, $\frac{\cos(\theta)}{\sqrt{\cos\theta\sin\theta}}$, say. What happens to that if you bring everything inside the square root.

I will get the following:
$\sqrt{\frac{cos^2\theta}{cos\theta sin\theta}}$ = $\sqrt{\frac{cos\theta}{ sin\theta}}$ = $\sqrt (\cot\theta)$

The problem is I'm not sure how to simplify:

$2\sqrt{\frac{R}{2sin\theta cos\theta}}\div\sqrt{g}\times(cos\theta-sin\theta)$

Should I bring everything into the $\frac{R}{2sin\theta cos\theta}$?

 

[haruspex]

The problem is I'm not sure how to simplify:
$2\sqrt{\frac{R}{2sin\theta cos\theta}}\div\sqrt{g}\times(cos\theta-sin\theta)$

Multiply out the parentheses, a(b-c)=ab-ac.

 

[jisbon]

Multiply out the parentheses, a(b-c)=ab-ac.

Multiplying will give me :

$2\sqrt{\frac{Rcos\theta}{2sin\theta cos\theta g}}-2\sqrt{\frac{Rsin\theta}{2sin\theta cos\theta g}} =2\sqrt{\frac{R}{2sin\theta g}}-2\sqrt{\frac{R}{2 cos\theta g}}=2(\sqrt{\frac{R}{2sin\theta g}}-\sqrt{\frac{R}{2 cos\theta g}})$

 

[haruspex]

Multiplying will give me :

$2\sqrt{\frac{Rcos\theta}{2sin\theta cos\theta g}}-2\sqrt{\frac{Rsin\theta}{2sin\theta cos\theta g}}$

No, you forgot to do something when you moved the trig factors inside the square root.

 

[jisbon]

Multiplying will give me :

$2\sqrt{\frac{Rcos\theta}{2sin\theta cos\theta g}}-2\sqrt{\frac{Rsin\theta}{2sin\theta cos\theta g}} =2\sqrt{\frac{R}{2sin\theta g}}-2\sqrt{\frac{R}{2 cos\theta g}}=2(\sqrt{\frac{R}{2sin\theta g}}-\sqrt{\frac{R}{2 cos\theta g}})$

Oops yep I forgot to square the first part, but is the second and third part correct?

 

[haruspex]

Oops yep I forgot to square the first part, but is the second and third part correct?

If you correct the first step the others change too, and you end up with the desired answer.

 

[jisbon]

If you correct the first step the others change too, and you end up with the desired answer.

$2\sqrt{\frac{Rcos^2\theta}{2sin\theta cos\theta g}}-2\sqrt{\frac{Rsin^2\theta}{2sin\theta cos\theta g}} =2\sqrt{\frac{Rcos\theta}{2sin\theta g}}-2\sqrt{\frac{Rsin\theta}{2 cos\theta g}}=2(\sqrt{\frac{Rcot\theta}{2 g}}-\sqrt{\frac{Rtan\theta}{2g}})$

 

[BvU]

Yep with square root r/2g in front

In your last posting you have $2\sqrt{R\over 2g} = \sqrt{2R\over g}$ in front !