Using energy conservation to find speed

[vu10758]

A stone of mass m is thrown straight up into the air with speed v_o. While the flight it feels a force of air resistance of magnitude f. Determine the speed of the stone when it hits the ground. You will need to combine energy conservation equations for the upward and downward parts of the motion.

This is what I did...

upward

(1/2)m(v_o)^2 - F = (1/2)(m)(v_f)^2 + mgh


downward

mgh - F = (1/2)m(v_o)^2

Am I correct so far? How do I combine them?

The correct answer is v = (v_o) SQRT(mg-f/mg+f). How do I get there?

 

[OlderDan]

You cannot add or subtract force and energy. You have to either equate forces with forces or energies with energies (work is like energy in that it is the mechanism by which energies are changed)

 

[vu10758]

Thanks for pointing that out
Now I have ...

upward

(1/2)m(v_o)^2 - F(h) = (1/2)(m)(v_f)^2 + mgh

downward

mgh - Fh = (1/2)m(v_o)^2

Is this correct?

 

[PhanthomJay]

Thanks for pointing that out
Now I have ...

upward

(1/2)m(v_o)^2 - F(h) = (1/2)(m)(v_f)^2 + mgh

downward

mgh - Fh = (1/2)m(v_o)^2

Is this correct?

That last term in the last equation...where you have v_o...that's not correct. You are trying to find its speed as it hits ground...

 

[OlderDan]

Thanks for pointing that out
Now I have ...

upward

(1/2)m(v_o)^2 - F(h) = (1/2)(m)(v_f)^2 + mgh

downward

mgh - Fh = (1/2)m(v_o)^2

Is this correct?

You need your first equation to find the maximum height. In that equation v_f is zero. Then use that h in the second equation, making the correction Jay noted.

 

[OlderDan]

I have a quick question. Wouldn't the v that the hitting the ground be the same as the v_o? Wouldn't the speed going up equals to the speed going down at ever position?

That would be true without the resistance, but not when the motion is constantly being opposed.

 

[vu10758]

Thanks for your help.

-Johnson