The unstretched rope is 20 meters. After getting dunked a few times the 80 kg jumper comes to rest above the water with the rope now stretched to 30 meters. What is the maximum length of the rope in meters when the jumper is being dunked?

Therefore maximum stretch is y2 = 32.36 m

Explanation:

In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium

        - W = 0

         k Δx = mg

         k = mg / Δx

         k = 80 9.8 / (30-20)

         k = 78.4 N / m

now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m

starting point. When will you jump

         Em₀ = U = mg y

final point. Just when the rope starts to stretch

          = K = ½ m v²

         Em₀ = Em_{f}

          mg y = ½ m v²

          v = √ 2g y

          v = √ (2 9.8 20)

          v = 19.8 m / s

now all kinetic energy is transformed into elastic energy

starting point

            Em₀ = K = ½ m v²

final point

            Em_{f} = + U = ½ k y² + m g y

            Emo = Em_{f}

           ½ m v² = ½ k y² + mgy

            k y² + 2 m g y - m v² = 0

we substitute the values and solve the quadratic equation

            78.4 y² + 2 80 9.8 y - 80 19.8² = 0

            78.4 y² + 1568 y - 31363.2 = 0

              y² + 20 y - 400 = 0

              y = [- 20 ±√ (20 2 +4 400)] / 2

              y = [-20 ± 44.72] / 2

the solutions are

              y₁ = 12.36 m

              y₂ = 32.36 m

These solutions correspond to the maximum stretch and its rebound.

Therefore maximum stretch is y2 = 32.36 m

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