created at April 24, 2021

Flying ball's trajectory

Consider folowing problem. Ball having a mass m and diameter d flies out from the starting point with a certain velocity v0 at an angle to the horizon α. Air's drag force is considered. Ball's rotation is not considered.

We will compose motion differential equations (in plane) and simulate ball's flying trajectory.

fig. 1 Forces acting on flying ball

Ball is impacted by gravity mg and air resistance force Fd (aka drag force), which acts opposite to the direction of velocity. Drag force is:

Fd=CdAρV22[N]Fd=Cd\cdot A \cdot \frac{\rho \cdot V^{2}}{2}\quad [N]


  • Cd - ball's drag coefficient
  • A - ball's frontal area
  • ρ - air density (1.29 kg/m3)
  • V - ball's current velocity

Cd depends on object's geometrical shape and is being detected experimentally. For ball shape let's consider Cd = 0.84.

Frontal Area is the one which will be observed when looking towards ball's motion direction. For ball obviously it makes no difference from where to look, so:

A=πd24A=\frac{\pi \cdot d^{2}}{4}

Finally we can write motion equations based on Newton's second law:

{mx¨=sign(x˙)CdAρx˙22my¨=sign(y˙)CdAρy˙22mg\begin{cases} m\cdot \ddot{x} = -sign(\dot{x})\cdot Cd\cdot A\cdot \frac{\rho \cdot \dot{x}^{2}}{2} \\m\cdot \ddot{y} = -sign(\dot{y})\cdot Cd\cdot A\cdot \frac{\rho \cdot \dot{y}^{2}}{2} - m\cdot g \end{cases}

We have to add -sign() factor since we need to take into account direction of velocity. Drag force should be opposite to motion direction, thus we put "minus" in there.

Now, if we solve this problem for: m = 0.4 kg, d = 0.22 m, v0 = 30 m/s, α = 30˚, the following trajectory will be obtained:

fig. 2 Ball's flying trajectory

Twisting motion of the ball will also affect the trajectory, however, this is not considered here.