I came across these ballistic equations to describe the motion of a point mass/projectile recently. Does anybody know how to derive the second equation, the change in angle with time?

$$\frac{dV}{dt} = -g \sin\theta -gkV^2 \tag{1}$$ $$\frac{d\theta}{dt} = -\frac{g \cos\theta}{V} \tag{2} $$ $$\frac{dx}{dt} = V \cos\theta \tag{3} $$ $$\frac{dy}{dt} = V \sin\theta \tag{4} $$

where $V$ is the velocity of the point mass, $\theta$ is the angle tangent to the trajectory of the point mass, $m$ is the mass of the particle, $x$ and $y$ are the Cartesian coordinates of the point mass, and $-gkV^2$ represents a drag force proportional to the square of the velocity.

  • $\begingroup$ what are your thoughts on how to? $\endgroup$ Apr 7, 2017 at 21:27
  • $\begingroup$ I understand that (3) and (4) are the resolved components of the velocity. Is it possible the coordinate system is defined in terms of tangential and normal coordinates rather than Cartesian coordinates? $\endgroup$ Apr 7, 2017 at 22:01
  • $\begingroup$ See the correct formulation of projectile motion with quadratic drag here. $\endgroup$ Apr 8, 2017 at 9:59
  • $\begingroup$ So in (1), the V is just the magnitude of the velocity, correct? But I still don't see where the angular dependence arises. Can you help there? $\endgroup$ Apr 8, 2017 at 14:13

1 Answer 1


Consider the cartesian coordinates of the mass $(x,y)$ and its velocity $$(\dot{x},\dot{y}) = V (\cos\theta,\sin\theta)$$

Differentiate the above (using the chain rule) to get the accelerations in an inertial coordinate frame

$$(\ddot{x},\ddot{y}) = \left( \dot{V} \cos\theta - V \dot{\theta}\sin \theta, \dot{V} \sin\theta + V \dot{\theta} \cos\theta \right) $$

and hence the equations of motion

$$(\ddot{x},\ddot{y}) = \left(-k V^2 \cos\theta, -g -k V^2 \sin\theta \right) $$

Equate the two above equations to get

$$\left. \begin{aligned} \dot{V} \cos\theta - V \dot{\theta}\sin \theta & = -k V^2 \cos\theta \\ \dot{V} \sin\theta + V \dot{\theta} \cos\theta & = -g -k V^2 \sin\theta \end{aligned} \right\} \begin{aligned} \dot{V} & = -g \sin\theta-k V^2 \\ V \dot{\theta} &= -g \cos\theta \end{aligned} $$

from which you get the equations in the OP.


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