My professor gave us no solutions and my final is in two days PLEASE HELP

  • Thread starter Thread starter ghettobacon
  • Start date Start date
  • Tags Tags
    Final Professor
Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem from a practice exam, with participants exploring the original poster's approach and understanding of the problem context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster shares their method for solving the projectile motion problem and seeks validation of their approach. Some participants emphasize the need for the original poster to show their work before receiving assistance.

Discussion Status

The discussion includes an exchange where the original poster attempts to clarify their reasoning and method. One participant confirms the correctness of the original poster's calculations, indicating a productive exchange of ideas.

Contextual Notes

The original poster expresses frustration over the lack of provided solutions from their professor, which adds pressure as their final exam approaches.

ghettobacon
Messages
2
Reaction score
0
I cannot for the love of god figure out to do this projectile motion problem, the first problem on this practice exam. Our professor for whatever reason decided not to give a solution page. Somebody PLEASE explain the first problem to me...and if youre feeling brave itd be nice if you could do a few of the others.

I uploaded the pdf to sharebeast, the link is here http://www.sharebeast.com/ozpiopxlfzg6

edit: didnt realize you can just add attachments on here, so i did
 

Attachments

Last edited by a moderator:
Physics news on Phys.org
Well, since this isn't "Do Your Homework For You.com", you'll need to actually show some work first. That is a simple projectile motion problem, you must have some idea on how to do it.
 
sorry I didnt imply it to be that, i just assumed I did this problem completely wrong.. well this is how I did it

method1:

i used rho= (v_final)^2/a_normal where rho is the radius of curvature and v_final would be at the top of the arc. plugging in, rho=7.5 m, a_normal=10 g/s^2 giving you v_final=sqrt(75)

applying v_final= v_o + at, in th x direction gives sqrt(75) = v_o*cos(30) which would mean v_o = 10 m/s

is this correct?
 
Yes, that's correct.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
Replies
2
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 99 ·
4
Replies
99
Views
22K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
3K