Single Variable Calculus, Chapter 5, 5.1, Section 5.1, Problem 14

The table below gives the velocity data for the shuttle between lift off and jettisoning of the solid rocket boosters. Use these data to estimate the height alone the earth's surface 62 seconds after lift off.

$\begin{array}{|c|c|c|c|c|c|c|}
\hline\\
\text{Event} & \text{Time} (s) & \text{Velocity}(\text{ft}/s)\\
\hline\\
\text{Launch} & 0 & 0\\
\text{Begin roll maneuver} & 10 & 185\\
\text{End roll maneuver} & 15 & 319\\
\text{Throttle to 89%} & 20 & 447\\
\text{Throttle to 67%} & 32 & 742\\
\text{Throttle to 104%} & 59 & 1325\\
\text{Maximum dynamic pressure} & 62 & 1445\\
\text{Solid rocket booster separation} & 125 & 4151\\
\hline
\end{array} $

We can find an upper estimate by using the final velocity for each time interval.


$
\begin{equation}
\begin{aligned}

D =& \sum \limits_{i = 1}^6 v(ti) \Delta ti
\\
\\
D =& (185) (10) + (319)(15 - 10) + (447)(20 - 15) + (742)(32 - 20) + (1325)(59 - 32) + (1445) (62 - 59)
\\
\\
D =& 54, 694 ft

\end{aligned}
\end{equation}
$