Hitting a rocket with a projectile

[MichaelTam]

Thanks, I finally got $\tan (\theta) = \frac {3A( A + 2 g )} {4 d B^2}$

 

[MichaelTam]

But...why it’s still incorrect?

 

[MichaelTam]

It is should be $\frac { 3A( A + 2g )} {4Bd}$?
Because $v_{x,0}= \frac {Bd(\sqrt{3A/B})} {3A}$
where,$v_{x,0} t_f= d$

 

[haruspex]

Thanks, I finally got $\tan (\theta) = \frac {3A( A + 2 g )} {4 d B^2}$

Use dimensional analysis to check your working.
Given the rocket's acceleration formula, A has dimension LT^-2, B is LT^-4. Your answer above fails the test.

 

[MichaelTam]

So the A must be in two dimension and B is 4 dimension, or A should be 1 dimension and B should be 3 dimension?

 

[MichaelTam]

 

[MichaelTam]

I had post all of m steps and assumptions, but I don’t know where I am wrong?

 

[MichaelTam]

Is this time I get the correct dimensions of both A and B?

 

[haruspex]

So the A must be in two dimension and B is 4 dimension, or A should be 1 dimension and B should be 3 dimension?

Sounds like you have not learned dimensional analysis. You should, it's quite easy and very useful.

 

[MichaelTam]

I know, but I find don’t know if 3Ag is in two dimension, and$4dB^2$ is in 4 dimensions respect to time?Can I count the g as a dimension for 3Ag?And is $4dB^2$ is not in 4 dimension but in 3 dimension?

 

[MichaelTam]

I just remember the analysis...when you tell me up about that tool.

 

[haruspex]

I know, but I find don’t know if 3Ag is in two dimension, and$4dB^2$ is in 4 dimensions respect to time?

[Ag]=LT^-2LT^-2=L²T^-4
[dB²]=L(LT^-4)²=L³T^-8.

 

[MichaelTam]

So it still incorrect?But where I am wrong...Is my assumption of $v_{x,0}$ is not correct or I make mistake on the $v_{y,0}$?

 

[haruspex]

So it still incorrect?But where I am wrong...

The point about using dimensional analysis is that you can check each step. Use 'binary chop'. Start in the middle of your working; if it's ok there go to three quarters of the way , etc.

 

[MichaelTam]

Ok, I am trying

 

[MichaelTam]

My assumption in the
$v_{y,0}-(gt_f/2)= (At_f/2) - (Bt_f/12)$ ( after simplify ) is not correct in dimension ?

 

[MichaelTam]

I still can’t find out how I get wrong although I know the thing I said before is wrong, I don’t know how to fix it...

 

[haruspex]

Your last line in post #36 is $\tan(\theta)=\frac{3A(A+2g)}{4Bd}$, which is dimensionally correct and looks like the right answer to me.
How do you know it is wrong?

 

[MichaelTam]

You mean plus or equal of the symbol?

 

[haruspex]

You mean plus or equal of the symbol?

=

 

[MichaelTam]

Thanks!

 

[MichaelTam]

I pass the course!,!,!

 

[MichaelTam]

Thanks!You a lot!Very much!

 

[haruspex]

I pass the course!,!,!View attachment 269324

Congratulations. (Do I get one too?)