Tarzan Swings
February 14th, 2008
Tarzan Swings
Physics Problem About Tarzan! Please Help.?
Tarzan swings on a 43.0 m long vine initially
inclined at an angle of 35.0? with the vertical.
The acceleration of gravity if 9.81 m/s2 .
What is his speed at the bottom of the
swing if he
a) starts from rest? Answer in units of m/s.
Its wrong. My teacher told us to use the PE+KE=PE+KE
Use energy equations.
K = 1/2 * m * v^2
P = m * g * h
Since he starts at rest, all of his energy is potential. To find his height, we take 43 (the length of the vine) and multiply it by the cosine of 35.0 degrees. P becomes
P = m * 9.81 * (43 * cos(35))
All of this will be converted into kinetic energy at the bottom of his swing (assume the bottom of the vine is level with the ground). Therefore,
P = K
m * 9.81 * (43 * cos(35)) = 1/2 * m * v^2
2 * 9.81 * (43 * cos(35)) = v^2
v = sqrt(2 * 9.81 * (43 * cos(35)))
v = 26.29 m/s
tarzan swing … Terri-Lynn screams
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