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March 6, 2015

March 6, 2015

Posted by **Ryan** on Thursday, September 3, 2009 at 6:06pm.

(a) Find the time of its fall.

(b) Find the height of its fall.

I've tried multiple times but I cannot get the correct answer.

- physics -
**bobpursley**, Thursday, September 3, 2009 at 6:10pmI assume that is .45m in the last second.

WEll, its average veloicty was .45m/s. Knowing its acceleration,

vf=vi+at in the last second, but avg velocity= vf+vi)/2 or

2 avgvel=vf+ vi or vi=2avgv-vf

vf=2avgv-vf+g

2vf=2avgv+g

vf=avgv + g/2 so solve for vf

Finally,

v=gtimefall solve for time of fall

h=1/2 g t^2

- physics -
**Ryan**, Thursday, September 3, 2009 at 6:20pmbobpursley,

No, the question actually says it travels 0.45h in the last second. So its 45% of the total height I believe. Unless it was a typo the average velocity is unknown. The only information known is its initial velocity (at the top of the fall, not only the last second) and the distance (0.45h) it falls in the last second.

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