Solution
The problem asks for the maximum length of the cord ($l$) that the jumper can use to ensure a safe clearance from the ground. We model this using the conservation of energy.
1. System Parameters:
Let the natural length of the cord used be $l$.
The stiffness of the available cord is given as $k = 33 \text{ N/m}$ for a length $l_0 = 30 \text{ m}$. Since the spring constant is inversely proportional to length, the stiffness $k’$ of the cut piece of length $l$ is: $$ k’ = \frac{k l_0}{l} $$ Substituting the given values ($k=33, l_0=30$): $$ k’ = \frac{990}{l} \text{ N/m} $$
Let the natural length of the cord used be $l$.
The stiffness of the available cord is given as $k = 33 \text{ N/m}$ for a length $l_0 = 30 \text{ m}$. Since the spring constant is inversely proportional to length, the stiffness $k’$ of the cut piece of length $l$ is: $$ k’ = \frac{k l_0}{l} $$ Substituting the given values ($k=33, l_0=30$): $$ k’ = \frac{990}{l} \text{ N/m} $$
2. Geometry of the Fall:
The jumper falls from the tower of height $H$. The cord is attached to the jumper’s feet. At the lowest point, the jumper’s head must be at a clearance $d$ from the ground.
The jumper falls from the tower of height $H$. The cord is attached to the jumper’s feet. At the lowest point, the jumper’s head must be at a clearance $d$ from the ground.
- Height of tower: $H = 62.5 \text{ m}$
- Height of jumper: $h = 1.5 \text{ m}$
- Clearance: $d = 1.0 \text{ m}$
3. Conservation of Energy:
The loss in gravitational potential energy of the jumper is converted into the elastic potential energy stored in the stretched cord. The center of mass (and the feet) descends by the total length of the cord plus extension ($l+\delta$). $$ m g (l + \delta) = \frac{1}{2} k’ \delta^2 $$ Substituting $(l+\delta) = 60$ and $\delta = (60-l)$: $$ m g (60) = \frac{1}{2} \left( \frac{990}{l} \right) (60 – l)^2 $$ Given $m = 66 \text{ kg}$ and $g = 10 \text{ m/s}^2$: $$ 66 \times 10 \times 60 = \frac{495}{l} (60 – l)^2 $$ $$ 39600 = \frac{495}{l} (3600 + l^2 – 120l) $$ Dividing both sides by 495: $$ \frac{39600}{495} = 80 $$ $$ 80 l = 3600 + l^2 – 120l $$ $$ l^2 – 200l + 3600 = 0 $$
The loss in gravitational potential energy of the jumper is converted into the elastic potential energy stored in the stretched cord. The center of mass (and the feet) descends by the total length of the cord plus extension ($l+\delta$). $$ m g (l + \delta) = \frac{1}{2} k’ \delta^2 $$ Substituting $(l+\delta) = 60$ and $\delta = (60-l)$: $$ m g (60) = \frac{1}{2} \left( \frac{990}{l} \right) (60 – l)^2 $$ Given $m = 66 \text{ kg}$ and $g = 10 \text{ m/s}^2$: $$ 66 \times 10 \times 60 = \frac{495}{l} (60 – l)^2 $$ $$ 39600 = \frac{495}{l} (3600 + l^2 – 120l) $$ Dividing both sides by 495: $$ \frac{39600}{495} = 80 $$ $$ 80 l = 3600 + l^2 – 120l $$ $$ l^2 – 200l + 3600 = 0 $$
4. Solving for l:
We solve the quadratic equation using the quadratic formula: $$ l = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$ $$ l = \frac{200 \pm \sqrt{(-200)^2 – 4(1)(3600)}}{2} $$ $$ l = \frac{200 \pm \sqrt{40000 – 14400}}{2} $$ $$ l = \frac{200 \pm \sqrt{25600}}{2} $$ $$ l = \frac{200 \pm 160}{2} $$ Two possible solutions: $$ l_1 = \frac{360}{2} = 180 \text{ m} \quad (\text{Rejected as } l > 60) $$ $$ l_2 = \frac{40}{2} = 20 \text{ m} $$ Therefore, the maximum length of the cord the jumper can use is **20 m**.
We solve the quadratic equation using the quadratic formula: $$ l = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$ $$ l = \frac{200 \pm \sqrt{(-200)^2 – 4(1)(3600)}}{2} $$ $$ l = \frac{200 \pm \sqrt{40000 – 14400}}{2} $$ $$ l = \frac{200 \pm \sqrt{25600}}{2} $$ $$ l = \frac{200 \pm 160}{2} $$ Two possible solutions: $$ l_1 = \frac{360}{2} = 180 \text{ m} \quad (\text{Rejected as } l > 60) $$ $$ l_2 = \frac{40}{2} = 20 \text{ m} $$ Therefore, the maximum length of the cord the jumper can use is **20 m**.