# Physics- Mechanical Energy

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:

Tom Mattson
Staff Emeritus
Gold Member
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.

But why though, I dont get it?

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

Double counting the Kinetic energy thing....

Tom Mattson
Staff Emeritus
Gold Member

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.