anonymous
  • anonymous
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 ? :/
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
please wait a moment, since I have to draw a table
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
I'm pondering....

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anonymous
  • anonymous
ok! :)
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
yes:)
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
so we have this situation: |dw:1433106277846:dw|
anonymous
  • anonymous
yes:)
Michele_Laino
  • Michele_Laino
at position P, we can write this equation (energy conservation): \[mg\frac{{3h}}{4} + \frac{1}{2}mv_0^2 = {E_{TOTAL}} = mgh\]
anonymous
  • anonymous
ok! what do we plug in?
Michele_Laino
  • Michele_Laino
dividing both sides by the product m*g, we get: \[\frac{{3h}}{4} + \frac{{v_0^2}}{g} = h\]
Michele_Laino
  • Michele_Laino
then we can write: \[\begin{gathered} \frac{{v_0^2}}{g} = h - \frac{{3h}}{4} \hfill \\ \hfill \\ \frac{{v_0^2}}{g} = \frac{h}{4} \hfill \\ \hfill \\ v_0^2 = \frac{{gh}}{4} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
how can we solve for that?
Michele_Laino
  • Michele_Laino
here is the next step: \[{v_0} = \sqrt {\frac{{gh}}{4}} \]
anonymous
  • anonymous
ok! what do we plug in?
anonymous
  • anonymous
would h be 24?
Michele_Laino
  • Michele_Laino
no, since we have: \[h = \frac{{{{24}^2}}}{{2 \times 9.81}} = ...meters\]
anonymous
  • anonymous
ohh okay so we get 29.357?
Michele_Laino
  • Michele_Laino
ok! h=29.4 meters
Michele_Laino
  • Michele_Laino
now we have: \[{v_0} = \sqrt {\frac{{gh}}{4}} = \sqrt {\frac{{9.81 \times 29.4}}{4}} = ...m/\sec \]
anonymous
  • anonymous
8.49?
Michele_Laino
  • Michele_Laino
yes! that's right!
Michele_Laino
  • Michele_Laino
so, what is the right option?
anonymous
  • anonymous
8.0 m/s?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
yay thanks!!
Michele_Laino
  • Michele_Laino
:)

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