Chapter 1.1, Problem 8P

Classical Mechanics. The differential equation for the velocity $v$ of an objects of mass $m$, restricted to vertical motion and subject only to the forces of gravity and air resistance, is

$m\frac{dv}{dt}=-mg-\gamma v$. (i)

In Eq. (i) we assume that the drag force, $-\gamma v$ is the drag force where $\gamma >0$ is drag coefficient is directly proportional to the velocity.

Acceleration due to gravity is denoted by $g$. Assume that upward direction is positive.

Show that the solution of Eq. (i) subject to the initial condition $v(0)=v_0$ is

$v=\left(v_0+\frac{mg}{\gamma }\right)e^{-\gamma t/m}-\frac{mg}{\gamma }$.

Sketch some integral curves, including the equilibriumsolution, for Eq. (i). Explain the physical significance of theequilibrium solution.

If a ball is initially thrown in the upward direction so that $v_0>0$, show that it reaches its maximum height when $t=t_{\max }=\frac{m}{\gamma }\ln \left(1+\frac{\gamma v_0}{mg}\right)$.

The terminal velocity of a baseball dropped from a hightower is measured to be $33\text{m/s}$. If mass of the baseball is $145\text{grams (g)}$ and $g\text{=}\text{9}\text{.8}{\text{m/s}}^2$, what is the value of $\gamma$?

Using the values of $m,g$, and $\gamma$ in part (d), what wouldbe the maximum height attained for a baseball thrown upward with an initial velocity $v_0=30\text{m/s}$ from a height of $2\text{m}$ above the ground?