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

AI Thread Summary
The discussion revolves around calculating the kinetic energy, initial velocity, and velocity at a specific height for a 0.5 kg ball thrown to a height of 20 meters. The kinetic energy is calculated using the formula KE = 1/2mv^2, with the final velocity at maximum height being 0 m/s. The user initially misunderstands the initial velocity but later correctly calculates the final velocity using the equation Vf^2 = 0 + 2(9.81)(20m), resulting in a velocity of 19.8 m/s. This leads to a kinetic energy of 98.1 Joules. The user questions whether the initial velocity is zero, indicating some confusion about the concepts involved.
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 ... ?

v_y^2=v_{0y}^2-2g\Delta y

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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