Chapter 5.2, Problem 84E

Bungee Jumping A bungee jumper plummets from a high bridge to the river below and then bounces back over and over again. At time t seconds after her jump, her height H (in meters) above the river is given by $H(t)=100+75e^{-t/20}\cos \left(\frac{\pi }{4}t\right)$. Find her height at the times indicated in the table.

t	H(t)
0
1
2
4
6
8
12

To determine

To find: The height of Bungee at the time indicated in the given table where the height above the river after her jump is given by $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$, where t is time in seconds measured after the jump.

Answer to Problem 84E

The height above the river at the different time is listed in below table.

$t$	$H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$
0	175
1	150.42
2	100
4	38.65
6	100
8	150.25
12	58.9

Explanation of Solution

The height above the river is $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$.

Substitute $t=0$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(0\right)=100+75e^{\left(\frac{0}{20}\right)}\cos \left(\frac{\pi }{4}\times 0\right)\\ =100+75e^0\cos \left(0\right)\\ =100+75\left(1\right)\left(1\right)\\ =175\end{array}$

So, $H\left(t\right)=175$ when $t=0$.

Substitute $t=1$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(1\right)=100+75e^{\left(\frac{-1}{20}\right)}\cos \left(\frac{\pi }{4}\right)\\ =100+75e^{\left(\frac{-1}{20}\right)}\left(0.707\right)\\ =100+75\left(0.951\right)\left(0.707\right)\\ =150.42\end{array}$

So, $H\left(t\right)=150.42$ when $t=1$.

Substitute $t=2$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(2\right)=100+75e^{\left(\frac{-2}{20}\right)}\cos \left(\frac{2\pi }{4}\right)\\ =100+75e^{\left(\frac{-1}{10}\right)}\cos \left(\frac{\pi }{2}\right)\\ =100+75\left(0.904\right)\left(0\right)\\ =100\end{array}$

So, $H\left(t\right)=100$ when $t=2$.

Substitute $t=4$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(4\right)=100+75e^{\left(\frac{-4}{20}\right)}\cos \left(\frac{4\pi }{4}\right)\\ =100+75e^{\left(\frac{-1}{5}\right)}\cos \left(\pi \right)\\ =100+75\left(0.818\right)\left(-1\right)\\ =38.65\end{array}$

So, $H\left(t\right)=38.65$ when $t=4$.

Substitute $t=6$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(6\right)=100+75e^{\left(\frac{-6}{20}\right)}\cos \left(\frac{6\pi }{4}\right)\\ =100+75e^{\left(\frac{-3}{10}\right)}\cos \left(\frac{3\pi }{2}\right)\\ =100+75e^{\left(\frac{-3}{10}\right)}\cos \left(\pi +\frac{\pi }{2}\right)\\ =100+75e^{\left(\frac{-3}{10}\right)}\cos \left(\frac{\pi }{2}\right)\\ =100\left[\because \cos \left(\frac{\pi }{2}\right)=0\right]\end{array}$

So, $H\left(t\right)=100$ when $t=6$.

Substitute $t=8$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(8\right)=100+75e^{\left(\frac{-8}{20}\right)}\cos \left(\frac{8\pi }{4}\right)\\ =100+75e^{\left(\frac{-2}{5}\right)}\cos \left(2\pi \right)\\ =100+75\left(0.670\right)\left(1\right)\\ =150.25\end{array}$

So, $H\left(t\right)=150.25$ when $t=8$.

Substitute $t=12$ in $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$,

$\begin{array}{c}H\left(12\right)=100+75e^{\left(\frac{-12}{20}\right)}\cos \left(\frac{12\pi }{4}\right)\\ =100+75e^{\left(\frac{-3}{5}\right)}\cos \left(3\pi \right)\\ =100+75\left(0.548\right)\left(-1\right)\\ =58.9\end{array}$

So, $H\left(t\right)=58.9$ when $t=12$.

Hence, the height above the river for $H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$ at indicated time is given in the below table.

$t$	$H\left(t\right)=100+75e^{\left(\frac{-t}{20}\right)}\cos \left(\frac{\pi t}{4}\right)$
0	175
1	150.42
2	100
4	38.65
6	100
8	150.25
12	58.9