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  
\[v_y =v^0_y\]
\[\frac{m}{k} ln(\frac{mg/k + v_y}{mg/k+v^0_y}) =-t \]

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

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

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