How to Model Projectile Motion with Differential Equations

Title: How to Model Projectile Motion with Differential Equations Category: physics

Set up the model

Projectile motion is a useful example of how to turn a physical situation into differential equations. The standard introductory model assumes a flat ground, constant gravitational acceleration, and no air resistance. Under those assumptions, the horizontal and vertical motions can be modeled separately and then combined to predict the trajectory.

Choose a horizontal coordinate $x$, with $x_0=0$ if the launch point is chosen as the horizontal origin, and let height $y$ be measured upward from the ground. Write the initial position as $(x_0,y_0)$ and the launch velocity as $\mathbf{v}_0$. If the launch speed is $v_0$ and its angle above the horizontal is $\theta$, its components are

$$v_{x}(0)=v_0\cos\theta, \qquad v_{y}(0)=v_0\sin\theta.$$

Newton’s second law is $m\mathbf{a}=\mathbf{F}$. With gravity as the only force, $\mathbf{F}=(0,-mg)$, so the mass cancels:

$$\frac{d^2x}{dt^2}=0, \qquad \frac{d^2y}{dt^2}=-g.$$

Near Earth’s surface, use $g\approx 9.81\,\text{m/s}^2$. The model’s initial conditions are $x(0)=x_0$, $y(0)=y_0$, and the velocity components above.

Solve the differential equations

Integrate each acceleration equation, applying the initial velocity and position. The resulting position functions are

$$x(t)=x_0+v_0\cos\theta\,t,$$ $$y(t)=y_0+v_0\sin\theta\,t-\frac{1}{2}gt^2.$$

These equations describe position while the projectile is in flight. For a launch with nonzero horizontal velocity ($v_0\cos\theta\ne0$), solve the first equation for time and substitute it into the second to get the trajectory as height versus horizontal distance:

$$y(x)=y_0+(x-x_0)\tan\theta-\frac{g(x-x_0)^2}{2v_0^2\cos^2\theta}.$$

For such a launch, the trajectory is a parabola in this model. For a vertical launch, $x$ remains constant instead. The equations also show why it helps to split the motion into components: horizontal velocity stays constant, while vertical velocity changes at a constant rate.

Worked example: find the landing distance

A ball is launched at $20\,\text{m/s}$ at $30^\circ$ above level ground from a height of $1.5\,\text{m}$. Take the launch point as $x_0=0$, use $g=9.81\,\text{m/s}^2$, and neglect air resistance. Its initial velocity components are

$$v_x=20\cos 30^\circ\approx17.32\,\text{m/s}, \qquad v_y=20\sin30^\circ=10\,\text{m/s}.$$

The height is therefore $y(t)=1.5+10t-4.905t^2$. The ball lands when its height is zero. Solving $1.5+10t-4.905t^2=0$ gives two roots; the positive one is

$$t=\frac{10+\sqrt{10^2+2(9.81)(1.5)}}{9.81}\approx2.18\,\text{s}.$$

The negative root does not describe a future landing. Since horizontal speed is constant, the landing distance from the launch point is

$$x(2.18)\approx17.32(2.18)\approx37.7\,\text{m}.$$

For another check, the ball reaches its maximum height when vertical velocity is zero. Since $v_y(t)=10-9.81t$, this occurs after about $1.02\,\text{s}$. Substituting that time into $y(t)$ gives a peak height of about $6.60\,\text{m}$ above the ground.

What changes when there is air resistance?

With drag, the acceleration depends on velocity, so the components are coupled through the drag force. A common idealization is linear drag, $\mathbf{F}_{\text{drag}}=-b\mathbf{v}$, where $b>0$ is a drag coefficient. Then

$$m\frac{dv_x}{dt}=-bv_x, \qquad m\frac{dv_y}{dt}=-mg-bv_y.$$

Quadratic drag, often used when drag is proportional to speed squared, has a force proportional to $-\lVert\mathbf{v}\rVert\mathbf{v}$. In either case, the simple parabolic trajectory generally no longer applies. Depending on the force law and desired accuracy, you can solve the equations analytically or use numerical methods. A realistic model also needs suitable drag parameters and may need to account for wind and changing air density.

Common modeling mistakes

• Mixing up signs: If upward is positive, gravity is $-g$ in the vertical equation. If you choose downward as positive, change the signs consistently.
• Using the wrong launch components: The angle is measured from the horizontal here. An angle measured from the vertical gives different component formulas.
• Treating time as the range: Flight time is in seconds; range is horizontal distance, found by evaluating $x(t)$ at the landing time.
• Using the wrong landing condition: If the projectile lands at ground level, set $y=0$. If it lands on a platform or at another elevation, set $y$ equal to that landing height.
• Assuming the ideal model is exact: The no-drag result is a baseline. At high speeds or over long flights, air resistance can make predictions differ substantially from observations.