Calculating Kinetic Energy, Velocity and Height for a .5kg Ball

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Homework Help Overview

The discussion revolves around calculating the kinetic energy, initial velocity, and velocity at a specific height for a 0.5 kg ball thrown upwards to a height of 20 meters. The subject area includes concepts from mechanics, specifically kinetic energy and motion under gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the kinetic energy formula and the relationship between velocity and height. There are attempts to clarify the meaning of variables in motion equations and to derive the final velocity at maximum height.

Discussion Status

Some participants have provided equations and calculations related to the problem, while others are questioning the definitions of variables and the assumptions made about initial conditions. There is a mix of interpretations regarding the initial velocity and the application of formulas.

Contextual Notes

Participants are navigating through the implications of using different equations of motion and the assumptions about the ball's velocity at the moment it is thrown. There is uncertainty about the initial velocity and how it relates to the kinetic energy calculation.

KatieLynn
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Homework Statement



If a .5kg ball is thrown up in the air and reaches a height of 20 meters.
A)What is the ball's Kinetic Energy?
B)What is its initial velocity, as it leaves the thrower's hand?
C)What is its velocity at 16 meters?


Homework Equations



KE=1/2mv^2


The Attempt at a Solution



Okay so for the first one I'm thinking I need to use KE=1/2mv^2, I have the mass but I need the velocity to solve for the kinetic energy. So then I thought maybe the velocity would just be 0 because it says "as it leaves the thrower's hand" as in the split second when its not moving?
 
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Use v^2=u^2+2as. At max height(which I believe to be 20m) the final speed=0
 
Last edited:
what do u and s stand for in that equation?
 
u=initial speed,s=displacement,v=final speed,a=acceleration
 
Shouldn't it be ... ?

[tex]v_y^2=v_{0y}^2-2g\Delta y[/tex]

LOL, I had this posted till you replied with that mgh thing, and since I don't know that much Physics I was like wow I'm totally off.
 
Okay here's what I did using that equation...

Vf^2= 0 + 2(9.81)(20m)
Vf=19.8m/s

then KE=1/2mv^2
so 1/2(.5kg)(19.8m/s)^2
KE=98.1

Correct?

Now would B just simply be 0?
 
Last edited:

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