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 $\mathbf{45}\mathbf{°}$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 $\mathbf{30}\mathbf{°}$ 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. Ignore air resistance and the mass of the vine.

The given data can be listed below as,

• The length of a end of a vine is, $L=20\mathrm{m}$.
• The initial angle of inclination with the vertical is, ${\theta }_1=45°$.
• The final angle of inclination with the vertical is, ${\theta }_2=30°$.

In this question, the energy conservation method concept can help determine Tarzan's speed just before he reaches Jane. The value of potential energy (gravitational) directly depends on the object's altitude above the ground.

Here, $h_1$ represents the initial height, $h_2$ represents the final height, $∆h$ represents the difference in height from the ground before and after the swing.

The expression to calculate the initial height $\left(h_1\right)$ is expressed as,

$h_1=L\left(1-\cos {\theta }_1\right)$

Substitute all the known values in the above expression.

$$\begin{gathered} h_1=\left(20\mathrm{m}\right)\left(1-\cos 45°\right) \\ \approx 5.86\mathrm{m} \end{gathered}$$

The expression to calculate the initial height $\left(h_2\right)$ is expressed as,

$h_2=L\left(1-\cos {\theta }_2\right)$

Substitute all the known values in the above expression.

$$\begin{gathered} h_2=\left(20\mathrm{m}\right)\left(1-\cos 30°\right) \\ =2.68\mathrm{m} \end{gathered}$$

The expression to calculate the difference in height from the ground before and after the swing is expressed as follows:

$∆h=\left(h_1-h_2\right)$

Here, $∆h$ is the difference in height from the ground before and after the swing.

Substitute all the known values in the above expression.

$$\begin{gathered} ∆h=\left(5.86\mathrm{m}-2.68\mathrm{m}\right) \\ =3.18\mathrm{m} \end{gathered}$$

The expression of the Tarzan’s speed just before he reaches Jane is expressed as,

$$\begin{gathered} KE=PE \\ \left(\frac{1}{2}m{v}^2\right)=\left(mg∆h\right) \\ \left(\frac{1}{2}{v}^2\right)=g∆h \\ v=\sqrt{2g∆h} \end{gathered}$$

Here, v is the Tarzan’s speed just before he reaches Jane, KE is the kinetic energy, PE is the potential energy, g is the gravitational acceleration whose value is$9.81\mathrm{m}/{\mathrm{s}}^2$ .

Substitute all the known values in the above equation.

$$\begin{gathered} v=\sqrt{2\left(9.18\mathrm{m}/{\mathrm{s}}^2\right)\left(3.18\mathrm{m}\right)} \\ \approx 7.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, With Tarzan going that fast, it is likely that he will knock Jane off.