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Posted by **mely** on Wednesday, January 26, 2011 at 1:02am.

What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 {m} wide and 9.00 {m} below the top of the cliff?

- physics -
**drwls**, Wednesday, January 26, 2011 at 5:41amCalculate the time to fall 9.00 meters, and call it T

T = sqrt (2H/g)= 1.355 s

Require that

V* T > 1.75 m

Vmin occurs when V = (1.75 m)/T

= 1.29 m/s

- physics -
**Henry**, Thursday, January 27, 2011 at 8:17pmmely, check your 1-26,12:52am post.

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