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

A daring 510 { N} swimmer dives off a cliff with a running horizontal leap, as shown in the figure below.

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 5:41am
Calculate 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 8:17pm
mely, check your 1-26,12:52am post.

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