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

Answer:

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

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:

$K_{B} = U_{g,A} - U_{g,B}$

$K_{B} = m \cdot g \cdot (h_{A}-h_{B})$

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g \cdot (h_{A}-h_{B})$

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

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:

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

$v\approx 7.896\,\frac{m}{s}$

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