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A rock that had been dropped from a high platform is moving with a kinetic energy of 180 J when it strikes the ground. Ignore air resistance. What kinetic energy did the rock have when it had fallen one-third of the distance to the ground? (Points : 1) 120 J 20 J 45 J 60 J

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use v^2 = 2as and KE = 0.5mv^2. When it hits the ground, \[v _{F}^2 = 2as\]1/3rd of the way to the ground, \[v^2 = 2a \frac{ s }{ 3} = \frac{ 1 }{ 3 } 2as = \frac{ 1 }{3 } v _{F}^{2}\] so notice that the velocity squared 1/3rd the way down is 1/3rd the velocity squared when it hits the ground. KE when it hits the ground: \[KE _{F} = 180= 0.5 m v _{F}^{2}\] KE 1/3rd the way down: \[KE = 0.5 m * (\frac{ 1 }{3 } v _{F}^{2}) = \frac{ 1 }{ 3 } 0.5m v _{F}^{2} = \frac{ 1 }{ 3 }KE _{F}\]
the correct answer is 120 j. there i added a drawing which shows the solution.
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@yunus your working is fine, but your answer is incorrect - the question asks "What kinetic energy did the rock have when it had fallen one-third of the distance to the ground?" 1/3rd of the distance TO the ground, not from the ground. You gave the kinetic energy when it has fallen 2/3rds of the distance to the ground.
I still gave you a medal for the method/drawing :)
Oh thanks. my English is not good thats why i couldnt understand it well. It still can be solved by using just what i did in the drawing instead of takin y value different. am i right?
if you take y=6.12/m then you find 59.976 joule which can be rounded to 60j.
Yep, your method works, as long as you use y/3 not 2y/3. Yours works the same as mine, just uses PE instead of velocity. Both give 60J

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