## Motion of a Projectile Subject to Air Resistance Proportional to Velocity

Suppose a projectile is moving subject only to the force of gravity and air resistance proportional to speed. Applying Newton's Second Law horizontally gives
$m \frac{dv_x}{dt}= -kv_x$

Separate variables to give
$\frac{dv_x}{v_x} = - \frac{k}{m} dt$
and integrate to give
$ln \frac{v_x}{v^0_x} = -\frac{kt}{m} \rightarrow v_x = v^0_x e^{-\frac{kt}{m}}$

Applying Newton's Second Law vertically gives
$m \frac{dv_y}{dt}= -kv_y - mg$

Separating variables gives
$m \frac{d v_y}{mg+kv_y} = - dt$

Integration, with
$v_y =v^0_y$
at
$t=0$
gives
$\frac{m}{k} ln(\frac{mg/k + v_y}{mg/k+v^0_y}) =-t$

Rearranging this expression for
$v_y$
gives
$v_y = (\frac{mg}{k} +v^0_y) e^{- \frac{kt}{m}} - \frac{mg}{k}$

At any point, the angle the velocity makes with the horizontal is
$\theta$
where
$tan \theta = \frac{v_y}{v_x} = \frac{(\frac{mg}{k} +v^0_y) e^{- \frac{kt}{m}} - \frac{mg}{k}}{v^0_x e^{-\frac{kt}{m}}}$