# EK and velocity question

1. Jun 3, 2009

### brentwoodbc

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

I thought maybe try and find the initial x/y components of the velocity but you cant find any angles, maybe the y component is found by intitial velocity is 0 at the top and it falls 4 metres? But the 4 metres above is not the max height. How do you do this?

Thank you.

2. Jun 3, 2009

### rock.freak667

Use conservation of energy. Use the top of the 15m as the reference line.

3. Jun 3, 2009

### brentwoodbc

So at "point p" Ep = 9.8x1.2x19=223J
at the start Ek= .5x1.2x16^2 = 153 ? Ep = 0

so E_total = Ep+ek

153= 223+Ek
Ek = - 70 ? What am I doing wrong?

4. Jun 3, 2009

### brentwoodbc

I know if I do etotal as Ek = 176 and Ep = 15x9.8x1.2 = 176
Etotal = 329

so at point p

329=223+ek
ek=106

which when rounded is 1.1x10^2 J which is the right answer, but in the reference frame where the ball starts there is no ep because its on the ground? right?

5. Jun 3, 2009

### rock.freak667

yes at that point Ep=0

6. Jun 3, 2009

### brentwoodbc

hmm....
then whats wrong here?

Last edited: Jun 3, 2009
7. Jun 3, 2009

### rock.freak667

Sorry there I though you choose the ground where the ball was a the reference point.

At point P, if the height is 19m, then you are choosing the actual ground (15m below the ball) as the reference line.

So the initial energy would be the kinetic energy AND gravitational potential energy (the ball is 15m above the reference point initially then)

8. Jun 3, 2009

### brentwoodbc

is it just total ek at start = 153
and it loses a certain amount to ep ......ep= 4x9.8x1.2 = 47
Et=ek+ep
153=ek+47
153-47=106

Ek =106joules I think.

9. Jun 3, 2009

### rock.freak667

If you are taking the lowest level as the reference line, then initially

Ei=mgh1+1/2mu2

$$E_i=(1.2)(9.81)(15)+\frac{1}{2}(1.2)(16)^2$$

And then at point P, Ei=mgh2+Ek.

10. Jun 3, 2009

thank you