posted by brittany .
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?
(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.
where did you get the 3.26 in part b? i don't understand that part really...i get the first part though
im sorry im having a off day, i see it now
wait what would the answer be to part b im not getting the right answer
Where did u get 5.84m