Projectile Motion

A projectile is any object that once launched or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

Projectile motion, also known as parabolic motion, is an example of composition of motion in two dimensions: an u.r.m. on the horizontal axis and a u.a.r.m. on the vertical axis. In this section, we will study:

• The concept and representation of parabolic motion
• Its equations
• The maximum height a body reaches in parabolic motion
• The time it is in the air
• The range
• The angle of the trajectory

Shall we begin?

Concept and representation

Projectile motion, also known as parabolic motion, consists in launching a body with a velocity that form an angle α with the horizontal. In the following figure, you can see a representation of the situation.

Equations

The equations for projectile motion are:

• The equations for the u.r.m. on the x-axis

  $$x=x_0+v_x·t$$​

• The equations for the u.a.r.m. on the y-axis

  $$v_y=v_{0y}+a_y·t$$

  $$y=y_0+v_{0y}·t+\frac{1}{2}·a_y·t^2$$

Since, as we said above, the velocity forms an angle α with the horizontal, the x and y components are determined using the most common trigonometric relationships:

Finally, taking into consideration the above said, that y₀ = H , x₀ = 0, and that a_y = -g , , we can rewrite the formulas as are shown in the following list. These are the final expressions for calculating kinematics magnitudes in projectile motion or parabolic motion:

• Position (m)

  • Horizontal axis

    $$x=v_x\cdot t=v_0·\cos \left(\alpha \right)·t$$

  • Vertical axis

    $$y=\mathrm{H}+{\mathrm{v}}_{0\mathrm{y}}·\mathrm{t}-\frac{1}{2}·g·t^2=\mathrm{H}+{\mathrm{v}}_0·\sin \left(\mathrm{\alpha }\right)·\mathrm{t}-\frac{1}{2}·g·t^2$$

• Velocity (m/s)

  • Horizontal axis

    $$v_x=v_{0x}=v_0·\cos \left(\alpha \right)$$

  • Vertical axis

    $$v_y=v_{0y}-g\cdot t=v_0·\sin \left(\alpha \right)-g\cdot t$$

• Acceleration (m/s²)

  • Horizontal axis

    $$a_x=0$$

  • Vertical axis

    $$a_y=-g$$

Equation of position and trajectory in projectile motion

The equation of position of a body helps us to know at what point it is at each instant of time. In the case of a body moving in two dimensions, remember that, generically, it is described by:

Substituting the above expressions of the position on the horizontal axis (u.r.m.) and on the vertical axis (u.a.r.m.) in the generic equation of position, we can get the expression of the equation of position for parabolic motion.

On the other hand, to know which trajectory the body follows, that is, its equation of trajectory, we can combine the above equations to eliminate t, getting:

$$\begin{gathered} y=H+v_{0y}·\left(\frac{x}{v_{0x}}\right)-\frac{1}{2}·g·(\frac{x}{v_{0x}}{)}^2=H+k_1·x-k_2·x^2 \\ {k}_1=\frac{v_{0y}}{v_x}; \\ {k}_2=\frac{1}{2·{v_{0x}}^2}·g \end{gathered}$$ 

As expected, this is the equation of a parabola.

On the other hand, frequently in exercises, you would be asked for some of the following values.

Maximum height

This value is reached when the velocity in the y-axis, v_y , is 0. Starting from the equation of velocity in the y-axis, and making v_y = 0, we get the time t that it takes the body in get to this height. From that time, and from the equations of position, we can calculate the distance to the origin in the both axes, the x-axis and y-axis.

Flight time

It is calculated for y = 0, the vertical component of the position. That is, the flight time is the time required for the height to become 0 (the projectile reaches the ground).

Range

It is the maximum horizontal distance, from the starting point of the motion to the point in which the body hits the ground. Once the strong>flight time is obtained, simply substitute in the equation of position of the horizontal component.

Angle of the trajectory

The angle of the trajectory in a given point is the same as the angle that the velocity vector form with the horizontal at that point. To calculate it, we get the components v_x and v_y and from the trigonometric definition of the tangent of an angle, we calculate α:

$$\tan \left(\alpha \right)=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{v_y}{v_x}\Rightarrow \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$$