Posted by **monic** on Wednesday, October 10, 2012 at 11:25pm.

Tarzan swings on a 34.0-m-long vine initially inclined at an angle of 33.0° with the vertical.

(a) What is his speed at the bottom of the swing if he starts from rest?

m/s

(b) What is his speed at the bottom of the swing if he pushes off with a speed of 5.00 m/s?

m/s

- physics -
**Elena**, Thursday, October 11, 2012 at 7:01pm
(a) PE= mgh=mgL(1-cosα),

KE= mv²/2.

PE=KE

mgL(1-cosα)= mv²/2,

v=sqrt{2gL(1-cosα)}= ….

(b) PE+KE1=KE

mgL(1-cosα) + mv₀²/2= mv²/2,

v = sqrt{2gL(1-cosα) + v₀²}= …

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