Solve Rocket Problem with Step-by-Step Guide | Imaginary Number Solution

Click For Summary

Homework Help Overview

The discussion revolves around a problem related to kinematics, specifically involving the motion of a rocket and the calculation of initial velocity using equations of motion. Participants are exploring the implications of their equations and the resulting values, including the occurrence of imaginary numbers in their calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to derive the initial velocity of a rocket using kinematic equations. There is a focus on identifying errors in the equations used, particularly regarding unit consistency and sign conventions. Some participants express confusion over the results, particularly when encountering imaginary numbers.

Discussion Status

There is an active exploration of the equations involved, with participants questioning the correctness of their formulations and discussing potential errors. Some guidance has been offered regarding the signs in the equations, and there is an acknowledgment of the need to reassess the assumptions made about the variables involved.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a specific focus on ensuring that the equations align with physical principles, particularly in terms of direction and magnitude of forces involved.

takelight2
Messages
12
Reaction score
1
Homework Statement
A toy rocket moving vertically upward passes by a 2.1 m -high window whose sill is 8.0 m above the ground. The rocket takes 0.13 s to travel the 2.1 m height of the window.


What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.
Relevant Equations
s = ut + 1/2at^2
This is what I tried and it makes perfect sense to me. When i plug in the numbers to what I ended up with, I get an imaginary number for U initial...

Work is attached below.
 

Attachments

  • EEBDAB96-74F2-41AA-82C4-D90668E3F106.jpeg
    EEBDAB96-74F2-41AA-82C4-D90668E3F106.jpeg
    97.9 KB · Views: 174
Physics news on Phys.org
One of the equations you wrote down, ##v_0 = u^2+2gh##, is incorrect. You can see that just from comparing the units on the lefthand side with the units of the righthand side.
 
vela said:
One of the equations you wrote down, ##v_0 = u^2+2gh##, is incorrect. You can see that just from comparing the units on the lefthand side with the units of the righthand side.
Oh yeah that makes sense. So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?
 
takelight2 said:
So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?
When you wrote the equation ##v_0 = u^2 + 2gh##, not only is there the mistake of not squaring ##v_0##, but there is also a sign error in the equation. Can you spot it?
 
takelight2 said:
Oh yeah that makes sense. So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?
It seems to me that, if you are assuming u to be the launching velocity and Vo the velocity of the rocket when it was by the top of the window, either both signs in that equation are incorrect, or the input value of g should be negative.
Since the rocket is decelerating while flying upwards, Vo < u.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot4

:cool:
 

Similar threads

  • · Replies 47 ·
2
Replies
47
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K