Question

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45∘ with the vertical, steps off his tree limb, and swings down and then up to Jane’s open arms. When he arrives, his vine makes an angle of 30∘ with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan’s speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

Answer 1

Tarzan will knock her off her limb. (CRASH!)

Step-by-step explanation:

The difference between a tender embrace or a knock out is derived of a convenient application of the Principle of Energy Conservation, in which a tender embrace mean the absence of kinetic energy and, most specific, speed, when Tarzan arrives to the tree, where Jane is waiting. Tarzan starts at rest. Hence:

`U_(g,A) = U_(g,B) + K_(B)`

The criteria is depicted as follows:

`K_(B) = U_(g,A) - U_(g,B)`

`K_(B) = m \cdot g \cdot (h_(A)-h_(B))`

`(1)/(2)\cdot m \cdot v^(2) = m \cdot g \cdot (h_(A)-h_(B))`

`v = \sqrt{2\cdot g \cdot (h_(A)-h_(B))}`

Heights are, respectively:

`h_(A) = 20\,m\cdot (1 - \cos 45^(\textdegree))`

`h_(A)\approx 5.858\,m`

`h_(B) = 20\,m\cdot (1 - \cos 30^(\textdegree))`

`h_(B) \approx 2.679\,m`

The final speed is:

`v = \sqrt{2\cdot (9.807\,(m)/(s^(2)) )\cdot (5.858\,m-2.679\,m)}`

`v\approx 7.896\,(m)/(s)`

Which mean that Tarzan will knock her off her limb. (CRASH!)