Let's break down the problem step by step: ### **Given Data:** - Width of river, \( d = 5.00 \) m - Length of vine, \( x = 8.90 \) m (the height above the opposite edge) - Initial angle with vertical, \( \theta_i = 60.0^\circ \) - Angle when vine breaks, \( \theta_f = 20.0^\circ \) - Tarzan starts from rest We need to **find the horizontal distance Tarzan travels after the vine breaks**. --- ## **Step 1: Find Tarzan's speed at the moment the vine breaks** Let Tarzan's mass be \( m \) (it will cancel out). **Energy Conservation:** - Initial potential energy = Final potential energy + Kinetic energy Let the lowest point (vertical) be the reference (potential energy = 0). ### **Initial Height (\(h_i\)):** \( h_i = x - x \cos \theta_i = x(1 - \cos \theta_i) \) ### **Final Height (\(h_f\)):** \( h_f = x - x \cos \theta_f = x(1 - \cos \theta_f) \) ### **Change in height:** \( \Delta h = h_i - h_f = x(\cos \theta_f - \cos \theta_i) \) **Potential energy lost = Kinetic energy gained:** \[ mgh = \frac{1}{2}mv^2 \] \[ mg[x(\cos \theta_f - \cos \theta_i)] = \frac{1}{2}mv^2 \] \[ 2g x (\cos \theta_f - \cos \theta_i) = v^2 \] \[ v = \sqrt{2g x (\cos \theta_f - \cos \theta_i)} \] Plug in the values: - \(g = 9.8\, \text{m/s}^2\) - \(x = 8.90\, \text{m}\) - \(\cos 60^\circ = 0.5\) - \(\cos 20^\circ \approx 0.940\) \[ v = \sqrt{2 \cdot 9.8 \cdot 8.90 \cdot (0.940 - 0.5)} \] \[ = \sqrt{2 \cdot 9.8 \cdot 8.90 \cdot 0.44} \] \[ = \sqrt{2 \cdot 9.8 \cdot 3.916} \] \[ = \sqrt{2 \cdot 38.3768} \] \[ = \sqrt{76.7536} \] \[ = 8.76\, \text{m/s} \] --- ## **Step 2: Find Tarzan's Position and Velocity Components When the Vine Breaks** Tarzan's position relative to the tree at the moment the vine breaks: - The horizontal distance from the tree: \( x \sin \theta_f \) - The vertical distance below the tree: \( x (1 - \cos \theta_f) \) But the tree is on the opposite edge, so Tarzan's horizontal distance from the starting edge: \[ d' = \text{width of river} + x \sin \theta_f \] ### **Velocity Components** At the moment the vine breaks, Tarzan's velocity is **tangential** to the circle: - Tangential velocity (\(v\)), at angle \( \theta_f \) - Direction: Perpendicular to radius (vine) The tangential direction makes an angle \( 90^\circ - \theta_f \) with the horizontal. The **horizontal component**: \[ v_x = v \sin \theta_f \] The **vertical component**: \[ v_y = v \cos \theta_f \] Plug in the values: - \(v = 8.76\, \text{m/s}\) - \(\sin 20^\circ \approx 0.342\) - \(\cos 20^\circ \approx 0.940\) \[ v_x = 8.76 \times 0.342 = 2.997 \approx 3.00\, \text{m/s} \] \[ v_y = 8.76 \times 0.940 = 8.234 \approx 8.23\, \text{m/s} \] --- ## **Step 3: Find Time to Hit the Ground** Tarzan's height above the ground when the vine breaks: \[ y = x (1 - \cos \theta_f) \] \[ y = 8.90 \times (1 - 0.940) = 8.90 \times 0.060 = 0.534\, \text{m} \] He falls a vertical distance of \( y = 0.534\, \text{m} \). Use the equation for vertical motion: \[ y = v_y t + \frac{1}{2}(-g)t^2 \] But since Tarzan is **above** the ground, and downward is negative: \[ 0 = y + v_y t - \frac{1}{2}gt^2 \] \[ 0 = 0.534 + 8.23 t - 4.9 t^2 \] Solve this quadratic for \( t \): \[ 4.9 t^2 - 8.23 t - 0.534 = 0 \] Use the quadratic formula: \[ t = \frac{8.23 \pm \sqrt{8.23^2 + 4 \times 4.9 \times 0.534}}{2 \times 4.9} \] \[ = \frac{8.23 \pm \sqrt{67.73 + 10.46}}{9.8} \] \[ = \frac{8.23 \pm \sqrt{78.19}}{9.8} \] \[ = \frac{8.23 \pm 8.85}{9.8} \] Take the positive root: \[ t = \frac{8.23 + 8.85}{9.8} = \frac{17.08}{9.8} = 1.74 \text{ s} \] --- ## **Step 4: Find the Horizontal Distance Traveled After the Vine Breaks** \[ \Delta x = v_x t = 3.00 \times 1.74 = 5.22\, \text{m} \] --- ## **Step 5: Check If He Makes It Across** He needs to land at least **5.00 m** from the starting edge. He starts from \( x \sin \theta_f \) from the tree **toward** the starting edge. But we want the total horizontal distance from the starting edge after Tarzan lets go: \[ \text{Total distance from starting edge} = x \sin \theta_f + \Delta x \] \[ = 8.90 \times 0.342 + 5.22 = 3.04 + 5.22 = 8.26\, \text{m} \] Since \( 8.26 > 5.00 \), **Tarzan safely lands on the other side.** --- ## **Final Answers** ### **a. Horizontal distance after the vine breaks:** \[ \boxed{5.22\, \text{m}} \] ### **b. Total horizontal distance from the starting edge:** \[ \boxed{8.26\, \text{m}} \] ### **c. Does he safely land?** \[ \boxed{\text{Yes, he safely lands on the other side.}} \]

When Tarzan's vine breaks, he continues moving horizontally with the velocity it had at that moment, determined by energy conservation. His horizontal displacement after breaking is calculated by multiplying this velocity component by the fall time, which is found from vertical motion equations considering gravity. Comparing the total horizontal distance traveled to the width of the river confirms whether he lands safely on the other side.