Hitting a rocket with a projectile

In summary: You need to put { and } around the subscripts on the v, otherwise only the y becomes a...In summary, the projectile hit the rocket just when the rocket reached its maximum height. The downward gravitational acceleration is 𝑔.
  • #36
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  • #37
I had post all of m steps and assumptions, but I don’t know where I am wrong?
 
  • #38
Is this time I get the correct dimensions of both A and B?
 
  • #39
MichaelTam said:
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.
 
  • #40
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?
 
  • #41
I just remember the analysis...when you tell me up about that tool.
 
  • #42
MichaelTam said:
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.
 
  • #43
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}##?
 
  • #44
MichaelTam said:
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.
 
  • #45
Ok, I am trying
 
  • #46
My assumption in the
## v_{y,0}-(gt_f/2)= (At_f/2) - (Bt_f/12)## ( after simplify ) is not correct in dimension ?
 
  • #47
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...
 
  • #48
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?
 
  • #49
You mean plus or equal of the symbol?
 
  • #50
MichaelTam said:
You mean plus or equal of the symbol?
=
 
  • #51
Thanks!
 
  • #52
I pass the course!,!,!
B06E2143-66C3-44D9-B030-158F265B1AD1.png
 
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  • #53
Thanks!You a lot!Very much!
 
  • #54
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