# Physics- Mechanical Energy

1. Apr 24, 2003

### PiRsq

An arrow is shot from ground (height=0) at a speed of 46m/s and travels in an archy line and at the maximum height has a speed of 42m/s. Arrows mass is, m...what is the maximum height that the arrow reaches? **IGNORE ANY FRICTION**

Thanks for any help guys

What I did was:

Et1=Mechanical energy before arrow was shot
Et2=Mechanical Energy after arrow was shot
Vf2=Final velocity squared
Vf1=Initial velocity squared

Et1=Et2

m(vf2)/2 - m(vi2)/2 = mgh + m(vf2)/2 - m(vi2)/2

...it doesnt work out though...why???

Last edited: Apr 24, 2003
2. Apr 24, 2003

### Tom Mattson

Staff Emeritus
You've got the right idea.

You've double-counted the kinetic energies. Each one should only appear once.

The total initial energy is Ei=(1/2)mvi2

The total final energy is Ef=(1/2)mvf2+mgh

Just equate and solve.

3. Apr 24, 2003

### PiRsq

But why though, I dont get it?

4. Apr 24, 2003

### Tom Mattson

Staff Emeritus
What don't you get? The expressions for the total energies, or the mistake of double counting kinetic energy?

5. Apr 24, 2003

### PiRsq

Double counting the Kinetic energy thing....

6. Apr 25, 2003

### Tom Mattson

Staff Emeritus
Re: Re: Physics- Mechanical Energy

Continuing with what I wrote earlier:

Equate the energies Ei=Ef

(1/2)mvi2=(1/2)mvf2+mgh

See? You put in an extra -(1/2)mvi2 on the RHS and an extra -(1/2)mvf2 on the LHS.