## Related questions with answers

Recall that the velocity of the free-falling bungee jumper can be computed analytically as:

$v(t)=\sqrt{\frac{g m}{c_{d}}} \tanh (\sqrt{\frac{g c_{d}}{m}} t)$

where v(t) = velocity (m/s), t = time (s), g = 9.81 m/s$^{2}$, m = mass (kg), $c_{d}$ = drag coefficient (kg/m). (a) Use Romberg integration to compute how far the jumper travels during the first 8 seconds of free fall given m = 80kg and $c_{d}$ = 0.2kg/m.Compute the answer to $\varepsilon_{s}=1 \%$. (b) Perform the same computation with quad.

Solution

Verified$\textbf{a.)}$ The distance traveled during the first 8 seconds can be calculated by finding the integral of the velocity from $t=0$ to $t=8$ s.

$\begin{aligned} d&=\int_{0}^{8}\sqrt{\frac{gm}{c_{d}}}\tanh{\left(\sqrt{\frac{gc_{d}}{m}}t \right)} dt\\ \end{aligned}$

The MATLAB script that applies the Romberg integration algorithm to evaluate this integral to an accuracy of $\epsilon_{s}=1\%$ is shown below.

```
clear;clc;close all;
m=80;cd=0.2;g=9.81;
f=@(t)sqrt(g*m/cd)*tanh(sqrt(g*cd/m)*t);
maxit=40; es=1; a=0;b=8;
n = 1;
I(1,1) = tra(f,a,b,n);
iter = 0;
while iter<maxit
iter = iter+1;
n = 2^iter;
I(iter+1,1) = tra(f,a,b,n);
for k = 2:iter+1
j = 2+iter-k;
I(j,k) = (4^(k-1)*I(j+1,k-1)-I(j,k-1))/(4^(k-1)-1);
end
ea = abs((I(1,iter+1)-I(2,iter))/I(1,iter+1))*100;
if ea<=es, break; end
end
d= I(1,iter+1);
```

*d* = 255.2417 m

**b.)** Using **quad()**

```
clear;clc;close all;
m=80;cd=0.2;g=9.81;
f=@(t)sqrt(g*m/cd)*tanh(sqrt(g*cd/m)*t);
d=quad(f,0,8);
```

$d=255.2600\mathrm{~ m}$

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