Determining Throwing Height: Dimensional Analysis

Click For Summary

Homework Help Overview

The discussion revolves around determining an expression for the maximum height a rock reaches when thrown straight up with an initial speed, using dimensional analysis. The problem involves variables such as speed, height, and gravity, while neglecting air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of a matrix for dimensional analysis, with some questioning the correctness of the reduction process. Others suggest an alternative approach by assuming powers for each variable to ensure dimensional correctness.

Discussion Status

The discussion is ongoing, with participants exploring different methods for dimensional analysis. Some guidance has been offered regarding the use of matrices versus direct power assumptions, but there is no explicit consensus on the correctness of the derived formula.

Contextual Notes

Participants note that the problem involves a limited number of variables, which may influence the choice of method for dimensional analysis. There is also a mention of a constant in the derived formula, which remains undefined in the discussion.

Firben
Messages
141
Reaction score
0
Throwing height:

A rock is thrown straight up with initial speed v. Determine a expression for the maximum height h the rock reach.The air resistance is neglected and the throwing height are dependent on the gravity.

Variable list:

Speed: V, LT^-1
Height h, L
gravity g, LT^-2

My matrix:

---L T--
V |1 -1|
L |1 0 |
g |1 -2|

After reduction i got

(k=constant)

k = h*g*v^2

It should be h = k*v^2/g

Any ideas ?
 
Physics news on Phys.org
I think you did the reduction wrong. The matrix is right though. It doesn't involve many variables, so you could do the problem without using a matrix.
 
Yes, but how?
 
Well, you can assume that each variable is raised to some power, and you know that the equation including them must be dimensionally correct, so then you can solve for what those powers are.

This is effectively the same as what you should be doing with the matrix. Using the matrix can make the answer easier to find when there are a lot of variables. But since there's only 3 variables, the matrix isn't really that useful.
 
yes. But is the formula right?
 
Firben said:
(k=constant)

k = h*g*v^2

It should be h = k*v^2/g

Any ideas ?

Their formula is right. (The one that 'it should be').
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
40
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K