Tarzan swings on a 25.0 m long vine initially inclined at an angle of 41.0° with the vertical.?

What is his speed at the bottom of the swing if he starts from rest.





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Tarzan swings on a 25.0 m long vine initially inclined at an angle of 41.0° with the vertical.?
Use conservation of energy.



25m(cos(41)) = 18.87 meters lower than the pivot (straight down).



So he's 6.13 meters higher than the lowest point in the swing.



Ug = mgh = m(9.8m/s^2)(6.13m) = (mass) x 60.1m2/s^2



At the bottom, all the potential energy(Ug) is now kinetic energy(K).



K = 1/2 mv^2



Ug = K



1/2 mv^2 = m60.1m2/s^2



cancel mass



1/2v^2 = 60.1m2/s^2



v^2 = 120.2m/s^2



v = 10.96m/s