Chapter 8.2, Problem 62ES

A bullet of mass m, fired straight up with an initial velocity of ${\upsilon }_0,$ is slowed by the force of gravity and a drag force of air resistance $k{\upsilon }^2,$ where k is a positive constant. As the bullet moves upward, its velocity $\upsilon$ satisfies the equation

$m\frac{d\upsilon }{dt}=-\left(k{\upsilon }^2+mg\right)$

where g is the constant acceleration due to gravity.

(a) Show that if $x=x\left(t\right)$ is the height of the bullet above the barrel opening at time t, then

$m\upsilon \frac{d\upsilon }{dx}=-\left(k{\upsilon }^2+mg\right)$

(b) Express x in terms of $\upsilon$ given that $x=0$ when $\upsilon ={\upsilon }_0.$

(c) Assuming that

$\begin{array}{l}{\upsilon }_0=988m/\text{s, }g=9.8m/{\text{s}}^{\text{2}}\\ m=3.56\times {10}^{-3}kg,k=7.3\times {10}^{-6}kg/m\end{array}$

use the result in part (b) to find out how high the bullet rises.