Tarzan swings on a 33.0 m long vine initially

Question

inclined at an angle of 31.0◦ with the vertical.
The acceleration of gravity if 9.81 m/s

What is his speed at the bottom of the
swing if he
a) starts from rest?
Answer in units of m/s
b) pushes off with a speed of 5.00 m/s?
Answer in units of m/s.
Thanks!

Answer 1

height above bottom call it h
h = 33(1-cos 31)

initial energy = m g h

speed at bottom = sqrt(2gh)

because
(1/2)mv^2 = m g h

for part b
initial energy = m g h +(1/2)m(25)
so at bottom
(1/2) m v^2 = that initial energy

Answer 2

Why you put 12(1-cos 31degrees)