Solving for Velocity: Common Mistakes and How to Get It Right Tonight

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Homework Help Overview

The discussion revolves around a physics problem related to projectile motion, specifically calculating the speed of a stone thrown from a building at an angle. Participants are attempting to understand the components of velocity and the application of kinematic equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the calculation of the x and y components of velocity, questioning the correctness of their values and the application of the speed formula. There is mention of potential rounding errors and the impact of initial conditions on the final speed calculation.

Discussion Status

Some participants have shared their calculations and results, noting discrepancies in their answers. There is an exploration of different methods, including the use of conservation of energy, with some guidance offered on alternative approaches. Multiple interpretations of the problem are being discussed, but no consensus has been reached.

Contextual Notes

Participants mention constraints such as the requirement for a specific accuracy in their answers and the challenges posed by the homework submission system. There are also comments on the educational environment and concerns about academic integrity.

NINHARDCOREFAN
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For number 11(please see the attachment), I keep messing up and I don't know what I'm doing wrong.

To find the answer I know I have to do this:
Speed = sqrt(Vxsquared + Vysquared)

I know that velocity of x is going to stay constant, so its v is 25.5

And for the velocity of y I did this(Vy=Vo*sin(ang)-gt):
Vy= 25.5*sin(29.1)-9.8(4.41) (time is right)
Vy= -30.82

so after plugging these values into the speed equation, I got this answer: 40.

What's wrong with this?
 
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For some reason, I can't attach the file. Here is the question.

Given: g = 9.8 m/s2
A stone is thrown from the top of a building upward at an angle of 29.1(degrees) to the horizontal and with an initial speed of 25.5 m/s. The height of the building is 40.7 m.

What is the speed of the stone just before it strikes the ground?
 
EDIT: ACK hold on... misread...
 
Last edited:
anyway I tried that too got 12.4 for Vy and 28.3 for speed which is still wrong
 
Last edited:
25.5m/s is the magnitude of the total velocity, therefore, it cannot be the same as the x component.
 
Damn it got it wrong again, I calculated Vx by 25.5*cos29.1 and still got the answer wrong. I only have 2 chances left to getting it right
 
What was the answer you got, NIN?
 
37.3 with my last calculation
 
That's a bit off my answer. Are you rounding off anywhere?

Vx = 25.5*cos(29.1) = 22.3
Vy = -30.8

sqrt(Vx^2+Vy^2)
 
  • #10
THANX A LOT.
 
  • #11
round off error?

plugged wrong number in?
 
  • #12
yep damn system accepts only 1% error wrong so 37.3 was wrong. 38 was right
 
  • #13
Is your prof doing anything to make sure that people aren't copying each others' work?

Submitting the HW to a website just seems like it's a quick easy way for unscrupulous students to cheat...
 
  • #14
LOL, last year the university was giving out homework by handouts, everybody cheated then. That's why they have HW on the web now. Everyone's numbers are different.
 
  • #15
Don't worry about x and y components: Use conservation of energy.

(1/2)mv_f^2=(1/2)mv_i^2+mgh

or

v_f=sqrt(v_i^2+2gh)=sqrt(25.5^2+2*9.8*40.7)=38.05
 
  • #16
krab,

I have a feeling that NIN hasn't gotten to that topic in his classes yet. It is still early in the semester.

You're right, though. Conservation is the way to go.
 

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