Posted by monic on Wednesday, October 10, 2012 at 11:25pm.
Tarzan swings on a 34.0mlong 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(1cosα),
KE= mv²/2.
PE=KE
mgL(1cosα)= mv²/2,
v=sqrt{2gL(1cosα)}= ….
(b) PE+KE1=KE
mgL(1cosα) + mv₀²/2= mv²/2,
v = sqrt{2gL(1cosα) + v₀²}= …
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