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 126,12:52am post.
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