Math Snap

STEP 1

What is this asking?
How high up is a bungee jumper when they're falling at a certain speed, given their weight, the height of the platform, and how stretchy the bungee cord is?
Watch out!
Don't forget that the bungee cord only starts stretching after the initial freefall!

STEP 2

1. Set up the energy equation
2. Solve for height

STEP 3

At the start, the jumper has only potential energy because they're not moving yet.
We can write this as $PE_i = m \cdot g \cdot h_i$, where $m$ is the mass ($54.5 \text{ kg}$), $g$ is the acceleration due to gravity ($9.8 \text{ m/s}^2$), and $h_i$ is the initial height ($85.0 \text{ m}$).

STEP 4

So, $PE_i = 54.5 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot 85.0 \text{ m} = 45351.5 \text{ J}$.
That's a whole lotta potential energy just waiting to be unleashed!

STEP 5

When the jumper is falling at $9.47 \text{ m/s}$, they have both potential and kinetic energy.
The potential energy is $PE_f = m \cdot g \cdot h_f$, where $h_f$ is the height we're trying to find.
The kinetic energy is $KE_f = \frac{1}{2} \cdot m \cdot v^2$, where $v$ is the velocity ($9.47 \text{ m/s}$).

STEP 6

Plugging in the numbers, $KE_f = \frac{1}{2} \cdot 54.5 \text{ kg} \cdot (9.47 \text{ m/s})^2 = 2440.08 \text{ J}$.
They're really zooming!

STEP 7

Don't forget the bungee cord!
When stretched, it stores elastic potential energy.
This energy is given by $EPE = \frac{1}{2} \cdot k \cdot x^2$, where $k$ is the spring constant ($58.5 \text{ N/m}$) and $x$ is how much the cord is stretched.
Since the cord only stretches after the first $10.1 \text{ m}$, the stretch is $x = 85.0 \text{ m} - h_f - 10.1 \text{ m} = 74.9 \text{ m} - h_f$.

STEP 8

Because energy is conserved, the total energy at the beginning equals the total energy at the end.
So, $PE_i = PE_f + KE_f + EPE$.
Let's plug everything in: $45351.5 \text{ J} = 54.5 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot h_f + 2440.08 \text{ J} + \frac{1}{2} \cdot 58.5 \text{ N/m} \cdot (74.9 \text{ m} - h_f)^2$.

STEP 9

Time to do some algebra!
Simplifying the equation, we get $42911.42 = 534.1 h_f + 29.25 (5610.01 - 149.8 h_f + h_f^2)$, which further simplifies to $29.25 h_f^2 - 4915.15 h_f + 122488.8 = 0$.

STEP 10

This is a quadratic equation in terms of $h_f$.
We can solve it using the quadratic formula: $h_f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 29.25$, $b = -4915.15$, and $c = 122488.8$.

STEP 11

Plugging the values into the quadratic formula, we get two possible solutions: $h_f = \frac{4915.15 \pm \sqrt{(-4915.15)^2 - 4 \cdot 29.25 \cdot 122488.8}}{2 \cdot 29.25}$.
This simplifies to $h_f \approx 66.5 \text{ m}$ and $h_f \approx 62.6 \text{ m}$.

STEP 12

Since the jumper is falling down, the height must be less than the initial height of $74.9 \text{ m}$ after the cord starts stretching.
Therefore, the correct answer is $h_f \approx 62.6 \text{ m}$ above the river.

The jumper is approximately 62.6 meters above the river when falling at $9.47 \text{ m/s}$.