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anonymous

  • one year ago

a rock dropped from a high platform is moving at 24 m/s downward when it strikes the ground. Ignore air resistance. How fast was the rock moving when it had fallen only one-fourth of the distance to the ground? 8.0 m/s , 6.0 m/s, 18.0 m/s , or 12.0 m/s ? :/

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  1. Michele_Laino
    • one year ago
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    please wait a moment, since I have to draw a table

  2. anonymous
    • one year ago
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    ok!

  3. Michele_Laino
    • one year ago
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    I'm pondering....

  4. anonymous
    • one year ago
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    ok! :)

  5. Michele_Laino
    • one year ago
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    we have the subsequent situation: |dw:1433105987038:dw| so total mechanical energy is: \[{E_{TOTAL}} = \frac{1}{2}m{v^2}\] where m is the mass of the ball or the mass of the rock

  6. anonymous
    • one year ago
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    yes:)

  7. Michele_Laino
    • one year ago
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    the mechanical energy, namely KE+PE has to be constant, during the free falling motion of our rock, and its value has to be equal to E_TOTAL

  8. Michele_Laino
    • one year ago
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    so we have this situation: |dw:1433106277846:dw|

  9. anonymous
    • one year ago
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    yes:)

  10. Michele_Laino
    • one year ago
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    at position P, we can write this equation (energy conservation): \[mg\frac{{3h}}{4} + \frac{1}{2}mv_0^2 = {E_{TOTAL}} = mgh\]

  11. anonymous
    • one year ago
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    ok! what do we plug in?

  12. Michele_Laino
    • one year ago
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    dividing both sides by the product m*g, we get: \[\frac{{3h}}{4} + \frac{{v_0^2}}{g} = h\]

  13. Michele_Laino
    • one year ago
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    then we can write: \[\begin{gathered} \frac{{v_0^2}}{g} = h - \frac{{3h}}{4} \hfill \\ \hfill \\ \frac{{v_0^2}}{g} = \frac{h}{4} \hfill \\ \hfill \\ v_0^2 = \frac{{gh}}{4} \hfill \\ \end{gathered} \]

  14. anonymous
    • one year ago
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    how can we solve for that?

  15. Michele_Laino
    • one year ago
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    here is the next step: \[{v_0} = \sqrt {\frac{{gh}}{4}} \]

  16. anonymous
    • one year ago
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    ok! what do we plug in?

  17. anonymous
    • one year ago
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    would h be 24?

  18. Michele_Laino
    • one year ago
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    no, since we have: \[h = \frac{{{{24}^2}}}{{2 \times 9.81}} = ...meters\]

  19. anonymous
    • one year ago
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    ohh okay so we get 29.357?

  20. Michele_Laino
    • one year ago
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    ok! h=29.4 meters

  21. Michele_Laino
    • one year ago
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    now we have: \[{v_0} = \sqrt {\frac{{gh}}{4}} = \sqrt {\frac{{9.81 \times 29.4}}{4}} = ...m/\sec \]

  22. anonymous
    • one year ago
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    8.49?

  23. Michele_Laino
    • one year ago
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    yes! that's right!

  24. Michele_Laino
    • one year ago
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    so, what is the right option?

  25. anonymous
    • one year ago
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    8.0 m/s?

  26. Michele_Laino
    • one year ago
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    that's right!

  27. anonymous
    • one year ago
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    yay thanks!!

  28. Michele_Laino
    • one year ago
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    :)

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