Posted by **Kelsey** on Wednesday, March 10, 2010 at 6:32pm.

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

- Physics -
**drwls**, Wednesday, March 10, 2010 at 6:56pm
(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.

- Physics -
**Kelsey**, Wednesday, March 10, 2010 at 7:34pm
Thank you SO much! I didn't realize i had to do 27(1-cos33)

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