Physics- Mechanical Energy

[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???

[Tom Mattson]

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

Et1=Et2


You've got the right idea.


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

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

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.

[PiRsq]

But why though, I dont get it?

[Tom Mattson]

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

[PiRsq]

Double counting the Kinetic energy thing....

[Tom Mattson]

Continuing with what I wrote earlier:

Originally posted by Tom
The total initial energy is Ei=(1/2)mvi2

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

Just equate and solve.

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.