College Algebra, Chapter 4, 4.6, Section 4.6, Problem 86

The equation $\displaystyle h(v) = \frac{Rv^2}{2gR - v^2}$ represents the maximum height $h$ (in meters) reached by the rocket that is fired upward from the surface of the earth with an initial velocity $v$ (in m/s). Where $R= 6.4 \times 10^6$ m is the radius of the earth and $g = 9.8$ m/s$^2$ is the acceleration due to gravity. Draw the graph of $h$ by using graphing device. Note that $h$ and $v$ must be positive. What does the vertical asymptote represent physically?

By substituting all the given values and by factoring

$\displaystyle h(v) = \frac{6.4 \times 10^6 (v^2)}{2 (9.8) (6.4 \times 10^6) - v^2} = \frac{6.4 \times 10^6 (v^2)}{125.44 \times 10^6 - v^2} = \frac{6.4 \times 10^6 (v^2)}{(11200 - v)(11200 + v)}$







It shows from the graph that the function has vertical asymptote at $v = 11200$ m/s. It means that the rocket will be at maximum height when the velocity of the rocket has reached $11200$ m/s. However, the rocket's velocity will eventually decrease and began falling.