- #1

- 96

- 0

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?

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- Thread starter vu10758
- Start date

- #1

- 96

- 0

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?

- #2

OlderDan

Science Advisor

Homework Helper

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- #3

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- 0

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?

- #4

PhanthomJay

Science Advisor

Homework Helper

Gold Member

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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....vu10758 said:

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?

- #5

OlderDan

Science Advisor

Homework Helper

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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.vu10758 said:

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?

- #6

OlderDan

Science Advisor

Homework Helper

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That would be true without the resistance, but not when the motion is constantly being opposed.vu10758 said: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?

- #7

- 96

- 0

Thanks for your help.

-Johnson

-Johnson

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