text: tarzan,,,,,,? tarzan swings on a 23.4 m long vine initially inclined at an angle of 32° 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 2.28 m/s? a. i agree with the answerers above about this problem involving the principle of conservation of energy.   at the point where tarzan starts his downward swing, he is 3.56m above the bottom of the swing. we get this by solving for the length of the base (adjacent side) of the triangle whose hypotenuse equals the length of the vine, i.e. 23.4m. the length of the base is equal to 23.4cos32=19.84m. deduct this from 23.4m and we get 3.56. therefore,  pe=mg*3.56     and  ke=1/2mv^2  equate the two based on our conservation of energy principle:  mg*3.56=1/2mv^2  (m cancels out)  v^2=69.77 v=8.35m/s  b)if he starts with an initial speed, then at that point where  he starts, he has ke in addition to pe.  ke at start=1/2m*2.28^2  ke at bottom of swing=1/2mv^2  pe+ke at start=ke at bottom  mg*3.56+1/2m*2.28^2=1/2mv^2  (m cancels out)  34.88+2.59=1/2v^2 v^2=37.47*2      =74.94 v=8.66m/s