A jonathan topliff

Jonathan Toplif 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 efects 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 efects of gravity and the projectile’s initial velocity, not the efects of other external forces.

Therefore, the horizontal velocity remains unchanged while the vertical velocity changes due to the efects of gravity. The parametric equations for a projectile’s position excluding air resistance can be seen below.

1 2 1 2 y=vtsin ( θ )− g t =v y t − g t 2 2

x=vtcos ( θ )=v x 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 efects 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 Resistanc e

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.

∑ F x =−k v x =m a x ∑ F x =−k v x −mg=ma y 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=

−k v x m

a y=

−k v y −g m

In order to determine the velocity and position, the equations must be solved as diferential equations.

First, the velocity and the position in the x axis will be determined.

ax=

d v x −k v x = dt m

dv x −k = dt vx m

S

d vx −k =S dt vx m

( ) ( )

e ln ⁡( v )=e x

−k

v x =e m

−k t m

t

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 v x0. 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

v x =v x 0 e m

t

−k

t m m s x =S ( v x ) = v x 0− v x 0 e m k k

Next, the velocity and the position in the y axis will be determined.

a y+

k v y d vy k vy = + =−g m dt m

ρ ( x ) =e

∫e ∫

kt m

(

S

( mk dt )

=e

kt m

kt m

)

kt

( ( )) e

kt

( )

dv k v y + y =−g∫ e m m dt

kt

e m v y=

dvy =−g∫ (e m ) dt kt

−gm m e +C k

−kt

−mg vy= +C e m k

The integral of the velocity is the position.

s y=

−mg m t− C e k k

−kt m

+C

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.

Ce

−kt m

=v y +

mg k

C e 0=C=v y 0+

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.

v −kt mg m (¿ ¿ y 0+ )e k −mg v y= +¿ k v v (¿ ¿ y 0+

mg ) k −kt

mg m )e +¿ k −mg m s y= t− ¿ k k

(¿ ¿ y 0+

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. Diferential 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 afected. 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 afected, resulting in a decreased maximum height.

v v (¿ ¿ y 0+

mg ) k −kt

mg m (¿ ¿ y 0+ )e +¿ k −mg m s y= t− ¿ k k −k

sx =

t m m v x 0− v x 0 e m k k