Math Problem Statement

Use bisection method to determine the drag coefficient needed so that a 95-kg bungee jumper has a velocity of 46 m/s after 9 s of free fall. Start with initial guesses of xl = 0.2 and xu = 0.5 and iterate until the approximate relative error falls below 5%.

Note: The acceleration of gravity is 9.81 m/s2. (Round the final answer to four decimal places.)

Solution

The bungee jumper’s velocity as a function of time is governed by the following equation:

v=gmc(1ecmt)v = \frac{gm}{c} (1 - e^{-\frac{c}{m}t})

where:

  • v=46v = 46 m/s (final velocity after 9s)
  • g=9.81g = 9.81 m/s² (acceleration due to gravity)
  • m=95m = 95 kg (mass of the jumper)
  • cc is the drag coefficient (what we are solving for)
  • t=9t = 9 s (time of fall)

The function to solve is:

f(c)=9.81×95c(1ec95×9)46f(c) = \frac{9.81 \times 95}{c} (1 - e^{-\frac{c}{95} \times 9}) - 46

We will use the bisection method to find the root of this function within the given range c[0.2,0.5]c \in [0.2, 0.5] and iterate until the approximate relative error is below 5%.

Now, I will perform the calculations. math

Given constants

g = 9.81 # m/s^2 m = 95 # kg v_target = 46 # m/s t = 9 # s

Function definition for the bisection method

def velocity_function(c): return (g * m / c) * (1 - math.exp(-c * t / m)) - v_target

Bisection method implementation

def bisection_method(xl, xu, tol): ea = 100 # Initial approximate error xr_old = xl iteration = 0

while ea > tol:
    iteration += 1
    xr = (xl + xu) / 2  # Midpoint
    f_xr = velocity_function(xr)
    f_xl = velocity_function(xl)

    # Check which side of the interval to replace
    if f_xl * f_xr < 0:
        xu = xr  # Root is in lower half
    else:
        xl = xr  # Root is in upper half

    # Compute approximate relative error
    if xr != 0:
        ea = abs((xr - xr_old) / xr) * 100

    xr_old = xr  # Update previous root estimate

return round(xr, 4), iteration

Initial bracket values

xl = 0.2 xu = 0.5 tol = 5 # 5% relative error

Compute root using bisection method

c_value, iterations = bisection_method(xl, xu, tol) c_value, iterationsThe required drag coefficient cc is 0.4813 (rounded to four decimal places). The bisection method converged in 4 iterations, achieving an approximate relative error below 5%.

Do you want a step-by-step breakdown of the iterations? Let me know if you have any questions!

Here are some related questions you might find interesting:

  1. How does the bisection method ensure convergence?
  2. What other numerical methods could be used to solve this problem?
  3. How would the solution change if the jumper's mass were different?
  4. What happens to the velocity equation if air resistance is neglected?
  5. How can we modify the approach to account for varying drag force with velocity?

Tip: The bisection method is a robust root-finding technique but can be slow compared to other numerical methods like Newton-Raphson, which has faster convergence under suitable conditions.

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Bisection Method
Exponential Decay
Physics of Free Fall

Formulas

v = (gm/c) (1 - e^(-c/m * t))
f(c) = (g * m / c) (1 - e^(-c * t / m)) - v_target

Theorems

Bisection Method

Suitable Grade Level

Grades 11-12 (Advanced Physics and Mathematics)