# Tarzan swings on a 30.0 m long vine initially inclined at an angle of 37.0º with the vertical. What is his sp?

A. Starts from Rest

You set up this problem by doing a drawing. Think about the vine hanging verically downwards.The end of the vine is 30m below the attachment point. Now the vine is moved sideways so that it makes an angle of 37° with the vertical. The end of the vine has now been lifted so that it is no longer 30m below the attachment point. From your drawing you should be able to see that the end of the vine is now
h = 30cos37° = 23.96m below the attachment point. If Tarzan clutches the vine it the end, he is 30-23.96m = 6.04m above the base level. He has potential energy which is equal to PE = m*g*h
When he swings down, as the vine passes the lowest base level, the kinetic energy is equal to Tarzan’s potential energy at the top of the swing. KE = 0.5*m*v²
Therefore: m*g*h = 0.5*m*v²
g*h = 0.5V²
9.81 * 6.04 = 0.5v²
V² = 59.254/0.5
V² = 118.5
v = 10.9m/s

For the second part of the problem if Tarzan has an itial speed of 4m/s, then he has the previous calculated potential energy, plus extra kinetic energy from his speed.
His total energy at the top is Pe = m*g*h + KE = 0.5*m*4²
This is all converted to KE at the bottom which is equal to the total energy at the top
m*g*h + 0.5*m*16 = 0.5*m*v²
Divide by m
9.81*6.04 + 0.5*16 = 0.5V²
59.25 +8/0.5 = v²
V² = 134.5
v = 11.6m/s which is speed at lowest point

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