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