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Physics 12 Online
OpenStudy (anonymous):

Want to make sure I've got the right idea with this problem (in comments)...

OpenStudy (anonymous):

If an object travels one half of its total path in the last second of its fall from rest. How high was it initially? \[x(t) = \frac{ 1 }{ 2 } g t^2\]\[x(t-1)=\frac{ 1 }{ 4 }[-g (t-1)^2]\]\[\frac{ -g t^2 }{ 2 }-\frac{ -g (t-1)^2 }{ 2 }=\frac{ -g t^2 }{ 4 }\] \[-\frac{ 1 }{ 2 }g(-2t+1)=-\frac{ g t^2 }{ 4 }\]\[\frac{ 1 }{ 4 }t^2-t+\frac{ 1 }{ 2 } = 0\]\[t^2-4t+2=0\]\[\frac{ 4\pm \sqrt{16-8} }{ 2 }\]\[t=3.41s, 0.5858s\]And the 0.5858s gets discarded because t>1. When I plug that in to the original x equation, I get\[-\frac{ 1 }{ 2 }(-9.81m/s^2)(3.41s)^2\]\[=57.036 m\] I looked this up to check my answer, and the number I saw was 48.something... I absolutely could not figure out how they got that number. Basically I just want to check my work... it's for an exam review and I'm stressing!

OpenStudy (anonymous):

Use the supposed answer to see if it is correct.

OpenStudy (anonymous):

It worked ^_^; I should have thought of that.

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