Posted by **brittany** on Thursday, April 15, 2010 at 3:56pm.

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 39° from 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 starts with an initial speed of 3.26 m/s?

- physics... -
**drwls**, Thursday, April 15, 2010 at 4:16pm
(a) Think consewervation of energy

M*g*H = (1/2) M V^2

V = sqrt(2 g H)

H is the elevation decrease.

H = 26.2 (1 - cos39) = 5.84 m

(b) M g H = CHANGE in kinetic energy

g*H = (1/2)[V^2 - 3.26^2]

Solve for V

This assumes that Tarzan's initial velocity is along the arc of motion of the swing. There could be some small change if it isn't.

- physics... -
**brittany**, Thursday, April 15, 2010 at 4:23pm
where did you get the 3.26 in part b? i dont understand that part really...i get the first part though

- physics... -
**brittany**, Thursday, April 15, 2010 at 4:26pm
im sorry im having a off day, i see it now

- physics... -
**drwls**, Thursday, April 15, 2010 at 4:29pm
Good

- physics... -
**brittany**, Thursday, April 15, 2010 at 4:29pm
wait what would the answer be to part b im not getting the right answer

- physics... -
**isaac**, Tuesday, April 20, 2010 at 3:24pm
Where did u get 5.84m

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