physics
posted by monic on .
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

(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}= …