# Projectile motion

1. Jan 17, 2008

### 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?

2. Jan 17, 2008

### rock.freak667

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

Last edited: Jan 17, 2008
3. Jan 17, 2008

### KatieLynn

what do u and s stand for in that equation?

4. Jan 17, 2008

### rock.freak667

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

5. Jan 17, 2008

### 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.

6. Jan 17, 2008

### 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?

Last edited: Jan 17, 2008