Asked by Michael
Tarzan, King of the Jungle (mass = 73.2 kg), grabs a vine of length 14.1 m hanging from a tree branch. The angle of the vine was 25.9° with respect to the vertical when he grabbed it. At the lowest point of his trajectory, he picks up Jane (mass = 41.0 kg) and continues his swinging motion. What angle relative to the vertical will the vine have when Tarzan and Jane reach the highest point of their trajectory?
Answers
Answered by
drwls
The grabbing of Jane at the lowest point of the swing is a momentum-conserving event, but some kinetic energy is lost. Then they swing to the highest point with energy conserved again.
Let Tarzan's speed before grabbing Jane = V1:
Tarzan's mass = Mt
Jane's mass = Mj
Mt*g*(1-cos25.9) = (Mt/2)*V1^2
V1 = sqrt[2*g*(1-cos25.9)] = 1.403 m/s
After grabbing Jane,
V1*Mt = V2(Mt+Mj)
V2 = V1*[Mt/(Mt+Mj)]= 0.641V1 = 0.899 m/s
Final maximum angle A is given by
g(1-cosA) = (1/2)V2^2
1 - cosA = 0.04126
cosA = 0.9587
A = 16.5 degrees
Let Tarzan's speed before grabbing Jane = V1:
Tarzan's mass = Mt
Jane's mass = Mj
Mt*g*(1-cos25.9) = (Mt/2)*V1^2
V1 = sqrt[2*g*(1-cos25.9)] = 1.403 m/s
After grabbing Jane,
V1*Mt = V2(Mt+Mj)
V2 = V1*[Mt/(Mt+Mj)]= 0.641V1 = 0.899 m/s
Final maximum angle A is given by
g(1-cosA) = (1/2)V2^2
1 - cosA = 0.04126
cosA = 0.9587
A = 16.5 degrees
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