Jane is running from the ivory hunters in the jungle. Cheetah throws a 7.0 -m-long vine toward her. Jane leaps onto the vine with a speed of $4.0 \mathrm{m} / \mathrm{s} .$ When she catches the vine, it makes an angle of \(20^{\circ}\) with respect to the vertical. (a) When Jane is at her lowest point, she has moved downward a distance \(h\) from the height where she originally caught the vine. Show that \(h\) is given by \(h=L-L \cos 20^{\circ},\) where \(L\) is the length of the vine. (b) How fast is Jane moving when she is at the lowest point in her swing? (c) How high can Jane swing above the lowest point in her swing?

We will first find the expression for the distance \(h\). At her lowest point, the vine will be vertical, and assuming the vine is always taut, we can use the cosine relationship to derive the value of \(h\). Let \(L\) be the length of the vine, and denote the angle between the vine and the vertical as \(\theta = 20^\circ\). The distance \(h\) can be determined as: $$ h = L - L\cos\theta $$ If we plug the given angle \(\theta = 20^\circ\) into the formula, we get: $$ h = L - L\cos 20^\circ $$ So the distance \(h\) is given by \(h = L - L\cos 20^{\circ}\).

To find out Jane's speed at the lowest point of the swing, we use the conservation of mechanical energy. The initial kinetic energy she has when she jumps onto the vine will be converted into gravitational potential energy and kinetic energy at the lowest point. Let the initial kinetic energy be \(E_{k,initial}\) and the final kinetic energy (when Jane is at the lowest point of her swing) be \(E_{k,final}\). The change in gravitational potential energy will be equal to the increase in kinetic energy: $$ mgh = E_{k,final} - E_{k,initial} $$ We can then rewrite the equation to solve for the final speed of Jane \(v_{final}\): $$ v_{final}^2 = v_{initial}^2 + 2gh $$ Given the initial speed \(4.0\ \mathrm{m/s}\) and that we derived the expression for \(h\), we can now plug in the values and solve for the final speed: $$ v_{final}^2 = (4.0\ \mathrm{m/s})^2 + 2(9.81\ \mathrm{m/s^2})(7.0\ \mathrm{m} - 7.0\ \mathrm{m}\cos 20^{\circ}) $$ Now calculate the final speed: $$ v_{final} = \sqrt{(4.0\ \mathrm{m/s})^2 + 2(9.81\ \mathrm{m/s^2})(7.0\ \mathrm{m} - 7.0\ \mathrm{m}\cos 20^{\circ})} $$ $$ v_{final} \approx 6.64\ \mathrm{m/s} $$ So Jane's speed at the lowest point in her swing is approximately \(6.64\ \mathrm{m/s}\).

To find the maximum height Jane can swing above the lowest point, we will again use the conservation of mechanical energy. At the highest point, all the kinetic energy will be converted into gravitational potential energy. We are given Jane's speed at the lowest point, which is \(6.64\ \mathrm{m/s}\). We can write the mechanical energy conservation equation as: $$ mgh_{max} = E_{k,lowest} - E_{k,highest} $$ Since the highest point has no kinetic energy, we can rewrite the equation as: $$ h_{max} = \frac{E_{k,lowest}}{mg} $$ Where \(E_{k,lowest} = \frac{1}{2}mv^2\). Now we can plug the values we know to find \(h_{max}\): $$ h_{max} = \frac{\frac{1}{2}(6.64\ \mathrm{m/s})^2}{9.81\ \mathrm{m/s^2}} $$ Finally, calculate the maximum height: $$ h_{max} \approx 2.24\ \mathrm{m} $$ So, Jane can swing approximately \(2.24\ \mathrm{m}\) above the lowest point in her swing.