Jonathan Topliff
MAT 266 Honors Contract
Projectile Motion with Air Resistance
The ideal representation of projectile motion is a parabola. Ideally, the projectile follows a perfectly
symmetrical path based solely upon its velocity and launch angle. However, situations and
circumstances are very rarely ideal. For instance, most projectiles on earth are not launched from a
vacuum. For this reason, the effects of air resistance must be accounted for in a realistic representation
of projectile motion.
First, aspects of ideal projectile motion will be explained to provide context. Like motion in general,
projectile motion can be split into several components. For the purposes of simplicity, it will be assumed
that the projectile will be traveling in only two dimensions, the x and y dimension. Therefore, the motion
of the projectile will be split into two components, which can be represented as parametric equations.
Keep in mind that these two components only include the effects of gravity and the projectile’s initial
velocity, not the effects of other external forces.
Therefore, the horizontal velocity remains unchanged while the vertical velocity changes due to the
effects of gravity. The parametric equations for a projectile’s position excluding air resistance can be
seen below.
1
1
𝑦 = 𝑣𝑡𝑠𝑖𝑛(𝜃) − 2 𝑔𝑡 2 = 𝑣𝑦 𝑡 − 2 𝑔𝑡 2
𝑥 = 𝑣𝑡𝑐𝑜𝑠(𝜃) = 𝑣𝑥 𝑡
As seen in the equations, the x position is steadily increasing, as there is not acceleration. However, the
y position is parabolic due to the influence of gravity.
When air resistance is incorporated into the equation, the motion of the projectile in both axes is
impacted. The effects of air resistance can be clearly seen via a free body diagram and the conservation
of energy.
y
Velocity
FARcos(θ)
)
x
FARsin(θ)
FAir Resistance
Weight
The force of air resistance acts in the opposite direction of velocity. The equation for air resistance is
listed below, where “k” is a drag constant that changes according to the shape of the object moving, and
“v” is the velocity of the object. The values for “k” usually vary according to surface area and similar
factors.
FAir Resistance = −𝑘𝑣
From this, and from the free body diagrams, a system can be set up in accordance with Newton’s
Second Law. The x and y components of the system are displayed below.
∑ 𝐹𝑥 = −𝑘𝑣𝑥 = 𝑚𝑎𝑥
∑ 𝐹𝑥 = −𝑘𝑣𝑥 − 𝑚𝑔 = 𝑚𝑎𝑦
By isolating for the acceleration, it allows for the determination of the equations for velocity and
position via integration. Acceleration is the derivative of velocity, and velocity is the derivative of
position.
𝑎𝑥 = −
𝑎𝑦 = −
𝑘𝑣𝑥
𝑚
𝑘𝑣𝑦
−𝑔
𝑚
In order to determine the velocity and position, the equations must be solved as differential equations.
First, the velocity and the position in the x axis will be determined.
𝑎𝑥 =
𝑑𝑣𝑥
𝑘𝑣𝑥
= −
𝑑𝑡
𝑚
𝑑𝑣𝑥
𝑘
= − 𝑑𝑡
𝑣𝑥
𝑚
𝑆(
𝑑𝑣𝑥
𝑘
) = 𝑆 (− 𝑑𝑡)
𝑣𝑥
𝑚
𝑘
𝑒 ln(𝑣𝑥 ) = 𝑒 −𝑚𝑡
𝑘
𝑣𝑥 = 𝑒 −𝑚𝑡
Assuming that vx will be zero at time=0, a complete equation for horizontal velocity can be created. This
initial condition results in a coefficient of vx0. The equation is listed below. To find the horizontal position
equation, the integral of the velocity must be taken. In addition, it must be assumed that the initial
position at time=0 is zero; this allows for the resulting constant to be solved. The horizontal position
equation is shown below.
𝑘
𝑣𝑥 = 𝑣𝑥0 𝑒 −𝑚𝑡
𝑠𝑥 = 𝑆(𝑣𝑥 ) =
𝑘
𝑚
𝑚
𝑣𝑥0 − 𝑣𝑥0 𝑒 −𝑚𝑡
𝑘
𝑘
Next, the velocity and the position in the y axis will be determined.
𝑎𝑦 +
𝑘𝑣𝑦 𝑑𝑣𝑦 𝑘𝑣𝑦
=
+
= −𝑔
𝑚
𝑑𝑡
𝑚
𝜌(𝑥) = 𝑒
𝑘𝑡
∫ 𝑒𝑚 (
𝑘
𝑆( 𝑑𝑡)
𝑚
𝑘𝑡
= 𝑒𝑚
𝑘𝑡
𝑑𝑣𝑦
𝑘
𝑣𝑦 +
) = −𝑔 ∫ (𝑒 𝑚 )
𝑚
𝑑𝑡
𝑘𝑡
𝑘𝑡
𝑑𝑣𝑦
)) = −𝑔 ∫(𝑒 𝑚 )
𝑑𝑡
𝑘𝑡
−𝑔𝑚 𝑘𝑡
𝑒𝑚 + 𝐶
𝑘
∫ (𝑒 𝑚 (
𝑒 𝑚 𝑣𝑦 =
𝑣𝑦 = −
𝑘𝑡
𝑚𝑔
+ 𝐶𝑒 − 𝑚
𝑘
The integral of the velocity is the position.
𝑠𝑦 = −
𝑘𝑡
𝑚𝑔
𝑚
𝑡 − 𝐶𝑒 − 𝑚 + 𝐶
𝑘
𝑘
However, the presence of constants in these equations requires further work. “C” can be defined by
determining the velocity at time 0, as shown below.
𝑘𝑡
𝐶𝑒 − 𝑚 = 𝑣𝑦 +
𝑚𝑔
𝑘
𝐶𝑒 0 = 𝐶 = 𝑣𝑦0 +
𝑚𝑔
𝑘
The constant “C” is then substituted into the two equations. The equations of projectile velocity and
position in the y axis are as follows.
𝑣𝑦 = −
𝑚𝑔
𝑚𝑔 −𝑘𝑡
+ (𝑣𝑦0 +
)𝑒 𝑚
𝑘
𝑘
𝑠𝑦 = −
𝑚𝑔
𝑚
𝑚𝑔 −𝑘𝑡
𝑚𝑔
𝑡 − (𝑣𝑦0 +
)𝑒 𝑚 + (𝑣𝑦0 +
)
𝑘
𝑘
𝑘
𝑘
In summation, projectile motion can be represented via parametric equations. By using a free body
diagram to determine the forces present in the x and y axis, a system can be set up that allows for
acceleration to be determined. Differential equations are then used to determine the velocity and
position equations for both the x and y axis. The position equations allow for the trajectory of the
projectile to be plotted. They are listed below. When compared to ideal projectile motion, it should be
noted that the motion on both the x and y axis are affected. There is now a negative acceleration value
along the x axis, which results in a shorter launch distance. The negative acceleration along the y axis
due to gravity is also affected, resulting in a decreased maximum height.
𝑠𝑦 = −
𝑚𝑔
𝑚
𝑚𝑔 −𝑘𝑡
𝑚𝑔
𝑡 − (𝑣𝑦0 +
)𝑒 𝑚 + (𝑣𝑦0 +
)
𝑘
𝑘
𝑘
𝑘
𝑠𝑥 =
𝑘
𝑚
𝑚
𝑣𝑥0 − 𝑣𝑥0 𝑒 −𝑚𝑡
𝑘
𝑘