How High Does an Arrow Reach if Shot at 46m/s?

[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 doesn't work out though...why?

 

[quantumdude]

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 doesn't work out though...why?

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

The total initial energy is E_i=(1/2)mv_i²

The total final energy is E_f=(1/2)mv_f²+mgh

Just equate and solve.

 

[PiRsq]

But why though, I don't get it?

 

[quantumdude]

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...

 

[quantumdude]

Continuing with what I wrote earlier:

Originally posted by Tom
The total initial energy is E_i=(1/2)mv_i²

The total final energy is E_f=(1/2)mv_f²+mgh

Just equate and solve.

Equate the energies E_i=E_f

(1/2)mv_i²=(1/2)mv_f²+mgh

See? You put in an extra -(1/2)mv_i² on the RHS and an extra -(1/2)mv_f² on the LHS.