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

Question

(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

Answer 1

(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}= …