Physics
posted by Kelsey .
Tarzan swings on a 27.0 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?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 3.00 m/s?
I tried to use energy equations and got the answer of final velocity = the square root of 2gh
But I don't think I am using the right thing for h.
please help

(a) If you use V = sqrt (2 g h), which is the correct formula, the h should be
27m *(1  cos33)
(b) Add M g h to the initial kinetic energy, and solve for the new V. The mass M cancels out.
V^2/2 = Vo^2/2 + g*h
Vo is the "pushoff speed", 3.00 m/s. 
Thank you SO much! I didn't realize i had to do 27(1cos33)