Question:(I) Jane, looking for Tarzan, is running at top speed\(\left( {{\bf{5}}{\bf{.0 m/s}}} \right)\)and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? Does the length of the vine affect your answer?

While running at top speed in the horizontal direction, Jane grabbed a vine hanging from a tree and achieved the maximum height when she swung upward.

According to the conservation of energy, the total energy before she grabbed the hanging vine equals the total energy after achieving the maximum height.

The energy before and after is the sum of the kinetic energy and potential energy.

Potential energy exists at the top, but there is no kinetic energy at the top. She runs along the horizontal, so there is kinetic energy at the reference level (ground) but no potential energy at the ground.

The given data can be listed below as:

• The initial speed of Jane is\({v_i} = 5.0{\rm{ m/s}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The diagram can be shown as:

Here, assume that the angle of inclination is very small\(\theta \), and can be neglected.

According to the conservation of energy, the energies can be written as:

\(\begin{array}{c}{E_i} = {E_f}\\P.{E_i} + K.{E_i} = P.{E_f} + K.{E_f}\\mg{h_i} + \frac{1}{2}mv_i^2 = mg{h_f} + \frac{1}{2}mv_f^2\end{array}\)

\(0 + \frac{1}{2}mv_i^2 = mg{h_f} + 0\) … (i)

Here,\({E_i}\)is the total energy at the ground before grabbing the vine, and\({E_f}\)is the total energy at the top after achieving some height.

The initial height\({h_i}\)is equal to zero,\({h_f}\)is the final height at the top,\({v_i}\)is the initial velocity at the ground, and\({v_f}\)is the final velocity at the top, which is equal to zero.

The initial potential energy \(P.{E_i}\) is equal to zero, \(K.{E_i}\) is the initial kinetic energy at the ground, \(P.{E_f}\)is the final potential energy at the top, and \(K.{E_f}\) is the final kinetic energy at the top, which is equal to zero.

From equation (i), the maximum height can be expressed as:

\(\frac{1}{2}mv_i^2 = mg{h_f}\)

\({h_f} = \frac{{v_i^2}}{{2g}}\) … (ii)

Substitute the values in the above expression.

\(\begin{array}{c}{h_f} = \frac{{{{\left( {5.0{\rm{ m/s}}} \right)}^2}}}{{2 \times 9.81{\rm{ m/}}{{\rm{s}}^2}}}\\ = 1.27{\rm{ m}}\end{array}\)

Thus, the maximum height she can swing upward is 1.27 m.

From equation (ii), it can be observed that height is not dependent on the vine’s length. So, the length of the vine does not affect the height.

Here, the length of the vine does not matter. However, the length of the vine should at least be 1.27 m long so that Jane can reach the maximum height upward.

From the above diagram, when the length of the vine is longer, the angle of inclination is smaller and vice versa. However, the height remains the same; that is, 1.27 m.

Thus, the length of the vine does not affect the answer.