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Physics- Mechanical Energy

  • Thread starter PiRsq
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  • #1
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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???
 
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Answers and Replies

  • #2
Tom Mattson
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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.
 
  • #3
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But why though, I dont get it?
 
  • #4
Tom Mattson
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What don't you get? The expressions for the total energies, or the mistake of double counting kinetic energy?
 
  • #5
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Double counting the Kinetic energy thing....
 
  • #6
Tom Mattson
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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.
 
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