projectile motion

[KatieLynn]

1. The problem statement, all variables and given/known data

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?


2. Relevant equations

KE=1/2mv^2


3. 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?

[rock.freak667]

Use v^2=u^2+2as. At max height(which I believe to be 20m) the final speed=0

[KatieLynn]

what do u and s stand for in that equation?

[rock.freak667]

u=initial speed,s=displacement,v=final speed,a=acceleration

[rocomath]

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.

[KatieLynn]

Okay heres 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?