Resistance is Futile Joseph Hays 16 May 2014 Page 1 of 24
Abstract When introductory physics books consider projectile motion, they often suggest that the reader assume air resistance is negligible; an assumption the same books will admit is rarely justified in reality. The purpose of this article is to empower the curious reader with a plainly spoken introduction to factoring air resistance into the motion of projectiles. In spite of the fact that the full range of projectile aerodynamics cannot be covered completely in this article, an engaged reader will nonetheless walk away armed with a basic differential equations tool-set which may then be expanded upon and applied toward producing more and more accurate models of a projectile s motion through air. Page 1 of 24
This article will focus on solving, graphing and comparing two models for projectile motion: a model where air resistance is neglected, and another model where air resistance is proportional to velocity. Both models will eventually be expressed as trajectory functions and compared to each other on the same plots. The profound difference between the two models will be demonstrated with several plots showing both trajectories together with varying mass and initial projection angle. Trajectory Through a Vacuum Before deriving an equation for trajectory with wind resistance, let s first recall the derivation of projectile trajectory through a vacuum from kinematic equations. Deriving this equation will serve as both a warm up for the strategy used to derive the trajectory equation which accounts for resistance, and also as a reference to which the air resistance model can be compared. The same general strategy will applied towards deriving both trajectory equations: First, displacement equations for each component (horizontal and vertical) of motion are derived. Next, the horizontal component is solved for time. Finally, that expression of time is substituted into the vertical displacement function to result in the desired trajectory equation. Page 1 of 24 Assume Air Resistance is Negligible Recall that a two dimensional vector may be broken up and analyzed in terms of its horizontal component and its vertical component. For a projectile traveling through two dimensions, its initial velocity vector, V 0, will have a horizontal component, V x0 = V 0 cos θ, and a vertical component, V y0 = V 0 sin θ, (see Figure 1). We may begin with the horizontal component.
Page 2 of 24 Figure 1: The initial velocity vector and its components.
Through a vacuum, there is no opposing force to cause a change in a launched projectile s horizontal velocity, so the horizontal position is simply: x = V x0 t (1) In the vertical direction, the acceleration of gravity in the only factor that will change the projectile s velocity, so the vertical position is: For trajectory, we first solve equation (1) for time: y = V y0 t 1 2 gt2 (2) t = x V x0 (3) Next, substitute this time value from equation (3) into equation (2) for trajectory: ( ) x Y vac = V y0 g ( x V x0 2 = V y0 x g x 2 V x0 2V 2 x0 V x0 ) 2 = V 0 sin θ V 0 cos θ x g 2V0 2 cos2 θ x2 = x tan θ gx 2 2V 2 0 cos2 θ Figure 2 shows a plot of the trajectory of a projectile launched into a vacuum at 60 meters per second at an angle 45 degrees from the horizontal. Note the impressive height and range that these conditions enable: about 92 and 360 meters, respectively. (4) Page 3 of 24
Page 4 of 24 Figure 2: Trajectory of a projectile launched at 60 m/s and 45 degrees from the horizontal.
Now that the trajectory equation for a projectile through a vacuum has been derived, the linear model will be examined. The linear model makes the fair assumption that air resistance is proportional to velocity. As before, we begin with the horizontal component. In contrast to the kinematics analysis before, however, setting up this differential equation will require an analysis of forces. Recall Newton s second law which explains that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass [2] often formalized as follows: F = m a (5) In the case of a projectile s horizontal component, the only force acting on it is the force air resistance, which means that equation (5) can be rewritten: ma x = bv x (6) The left side of equation (6) expresses the net force acting on the projectile (ma x ), and the right side expresses the elements that add up to the net force (in this case, only air resistance). By expressing acceleration (a x ) as the time derivative of velocity, equation (6) becomes as follows: Page 5 of 24 m dv x = bv x (7) dt Dividing both sides of equation (7) by mass (m) expresses a differential equation in terms of acceleration: dv x dt = bv x m, V x(0) = V x0 (8)
The left side of equation (8) expresses horizontal acceleration as the derivative of velocity (V x ) with respect to time. On the right side of equation (8), the negative sign expresses that the force of air resistance is opposite the direction of velocity, b is a constant of proportionality, and mass (m) in the denominator is expressed as an inverse proportion to the acceleration on the left hand side. Equation (8), as it is, can be used to describe the horizontal motion of a projectile. On the other hand; in order to describe the vertical motion, gravity (g) must be taken into account - this is a simple accommodation: Fy = ma y ma y = mg bv y m dv y = mg bv y (9) dt Dividing both sides of equation (9) by mass gives us an equation expressing acceleration in the vertical direction: dv y dt where g is the acceleration of gravity. In sum, our model so far can be expressed in components as follows: Horizontal Motion: Vertical Motion: dv x dt = g bv y m, V y(0) = V y0 (10) = bv x m, V x(0) = V x0 (11) dv y = g bv y dt m, V y(0) = V y0 (12) where θ is the projection angle, V x0 = V 0 cos(θ), and V y0 = V 0 sin(θ). The goal now is to solve these differential equations for velocity, which is a function of time. Page 6 of 24
Solutions In this section, the equations for horizontal and vertical motion will be solved. The solution for the projectile s horizontal position will be solved for time, then substituted into the solution for the projectile s vertical position to produce an equation for trajectory. The linear model s trajectory will then be plotted on the same graph as the model neglecting air resistance and the two will be compared for varying m, V 0, and θ. Horizontal Velocity Recall equation (11): dv x = bv x dt m, V x(0) = V x0 By solving this differential equation for V x we derive a formula for the horizontal velocity of the projectile. Begin by separating the variables multiply both sides by m/b: Now multiply both sides by dt: m dv x = V x b dt m b dv x = V x dt After dividing both sides by V x, we are ready to integrate: m dv x = dt b V x For ease of computation, definite integrals will be evaluated: Page 7 of 24
m b Vx V x0 dv x V x = t 0 dt (13) Notice that on the left-hand side of equation (13), the limits of integration span from the initial velocity value, V x (0) = V x0, to some arbitrary velocity V x. On the right hand side, the limits of integration span from the initial time value t = 0 to an arbitrary time t. Now, to evaluate the integrals in equation (13): Vx m dv x b V x0 V x = [ m b x ] ln V Vx Multiply both sides of equation (14) by b/m: Exponentiate both sides of equation (15): t 0 dt V x0 = t (14) ln V x Vx V x0 = b m t ln V x ln V x0 = b m t ln V x = b m t (15) V x0 Page 8 of 24 V x V x0 = e bt/m (16)
Finally, multiply both sides of equation (16) by V x0 for our solution to horizontal velocity: V x = V x0 e bt/m (17) Figure 3 gives us a feel for the behavior of equation (17). As we might expect, horizontal velocity decays exponentially towards zero for a 1 kg projectile launched at 60 meters per second, 45 degrees from the horizontal, where b = 1 (more on b later), the projectile loses almost all of its horizontal motion after only about 6 seconds. Horizontal Position First, recall that velocity is the time derivative of position. Therefore, in order to find the formula for the projectile s horizontal position, we must integrate equation (17): x = t 0 V x0 e bt /m dt t = V x0 e bt /m dt 0 = m b V x 0 e bt /m t 0 = mv x0 ( e bt/m 1 ) b Finally, the equation for horizontal position: x = mv x0 ( 1 e bt/m ) (18) b Page 9 of 24
Page 10 of 24 Figure 3: Horizontal velocity of a 1 kg projectile launched at 60 m/s and 45 degrees from the horizontal, b = 1.
Page 11 of 24 Figure 4: Horizontal position of a 1 kg projectile launched at 60 m/s and 45 degrees from the horizontal, b = 1.
Figure 4 gives us a sense of how far the same projectile from Figure 3 will travel over time. Recall that velocity for this particular projectile decays to essentially zero after about 6 seconds the graph in Figure 4 seems to agree with Figure 3, and the projectile travels 42 meters in the horizontal after about 6 seconds. Vertical Velocity We will now find a formula for the vertical velocity of the projectile - recall equation (12): dv y dt = g bv y m, V y(0) = V y0 Equation (12) is separable. Begin by multiplying both sides by dt and factoring the minus sign out of the right-hand side. ( dv y = g + bv ) y dt m Now divide both sides by (g + bv y /m): dv y g + b m V y Equation (19) is now ready for integration: = dt (19) Page 12 of 24
Vy t dv y V y0 g + b = dt m V y 0 [ m b ln g + b ] m V Vy y = t V y0 [ ln g + b ] m V Vy y = b V y0 m t ln g + b m V y ln g + bv y 0 m = b m t g + b ln m V y g + bv y 0 = b m t m g + b m V y g + bv y 0 m = e bt/m Solving for V y from here will show that: ( mg ) V y = b + V y0 e bt mg m (20) b Figure 5 shows us the vertical velocity of the same projectile from previous figures. Notice that the projectile s vertical velocity crosses from positive to negative after only about 2 seconds and then falls down with decreasing velocity until around 7-10 seconds where its velocity bottoms out, this velocity, about -10 m/s, is the terminal velocity. Page 13 of 24
Page 14 of 24 Figure 5: Vertical velocity of a 1 kg projectile launched at 60 m/s and 45 degrees from the horizontal, b = 1.
Vertical Position We now integrate equation (20) into a formula for the projectile s vertical position: t ( ( mg y = 0 b + V ) y 0 e bt /m mg ) dt b ( mg ) t = b + V y 0 e bt /m dt mg t dt 0 b 0 = m ( mg ) ( ) b b + V y 0 e bt/m 1 mg b t = m ( mg ) ( b b + V y 0 1 e bt/m) mg b t (21) A possible simplification for equation (21) is as follows: ( gm 2 ) + bmv ( y0 y = b 2 1 e bt/m) mg b t (22) Figure 6 shows the vertical position over time. Again the conditions are the same as the previous plots. We can now see that the position reaches a maximum around 1.7 seconds, reaches zero after about 5.2 seconds, and decreases seemingly linearly beyond that point. Trajectory through Air Recall equation (18): x = mv x0 ( 1 e bt/m ) b Page 15 of 24
Page 16 of 24 Figure 6: Vertical position of a 1 kg projectile launched at 60 m/s and 45 degrees from the horizontal, b = 1.
We will now derive a formula for projectile trajectory through air. The process is analogous to the process for finding trajectory through a vacuum we must first solve equation (18) for time (t); doing so, we produce the following: t = m ( ) b ln mvx0 (23) mv x0 bx Substituting this value of t into equation (22) and simplifying gives us a formula for trajectory: ( m 2 g + bmv y0 Y = b 2 ( m 2 g + bmv y0 = b 2 ) 1 e b/m ) 1 e ln ( m 2 g + bmv y0 = b 2 ( m 2 ) ( g + bmv y0 bx = = b 2 ( mg bv x0 + V y 0 V x0 m mv b ln x 0 mv x0 bx mv x 0 bx mv x0 ) ( ( 1 1 bx mv x0 ) x + m2 g b 2 ln mv x0 ) + m2 g b 2 ( 1 b mv x0 x )) + m2 g mg b ( ) + m2 g mvx0 bx ln b 2 ( ln 1 bx b 2 mv x0 ( ln 1 bx mv x0 [ ( )] m b ln mvx0 mv x0 bx mv x0 But remember that V x0 = V 0 cos θ, and V y0 = V 0 sin θ. Substituting these back into equation (24) and simplifying gives us our final result: ) ) ) (24) Page 17 of 24
( mg sec θ Y = bv 0 ) ( + tan θ x + m2 g b 2 ln 1 b sec θ ) x mv 0 We can finally make a comparison between the two models, but first, we need to use realistic values of mass and the proportionality constant b. But how do we find b? One way is through experimentation: Let s consider, for instance, a certain paintball with a mass of 3.2 grams that was tested to have a terminal velocity of 21.8 meters per second. b may be found by first reconsidering the model for vertical motion: dv y = g bv y (26) dt m An object dropped from rest will accelerate until it reaches its terminal velocity, at which point the acceleration will equal zero. Under these conditions, and letting the y-axis point downward, equation (26) can be written: Solving equation (27) for b, shows: 0 = g bv term m b = (25) (27) mg V term (28) Substituting the paintball s mass and terminal velocity into equation (28) shows that b, in this case, is 0.00144kg/s. We may now plot this paintball s trajectory and compare it to the model that neglects air resistance. As Figure 7 reveals, the difference between the two models for projectile motion is profound. Now, observe what happens when the projection angle is changed to 24 in Figure. Notice in Figure that when the projection angle is changed from 45 to 24, the range through a vacuum decreases as expected, but the range through air increases. Recall Page 18 of 24
Page 19 of 24 Figure 7: Trajectories neglecting and accounting for air resistance for a 3.2 gram paintball launched at 90 m/s (about maximum tournament velocity), 45 degrees from the horizontal. b = 0.00144kg/s
Page 20 of 24 Figure 8: A comparison of trajectories resulting from different projection angles. Max range without air is attained with a projection angle of 45. Max range through air is attained with a projection angle of 24.5 ± 0.5
that through a vacuum, the projection angle that will result in maximum range is 45. Through air, the angle that will result in maximum range is 24.5 ± 0.5. Figure 9 shows an interesting outcome from increasing the mass of the projectile while holding all other variables constant. Notice that as mass increases, the trajectory through air more closely resembles the trajectory through a vacuum. Do not, however, assume that a relatively higher mass implies that air resistance is negligible. Recall the expression for b in equation (28): b = mg V term This expression for b shows that for constant V term and g, b will increase directly with increasing mass. This observation calls for the trajectory examination of an object of higher mass: a bowling ball. Take, for instance, a 7.62kg bowling ball with a terminal velocity of 83.1 m/s. The resulting value of b is 0.8995 kg/s. As Figure 10 shows, the difference in range, even when the projection angle maximizes range through air, is over 200 meters. We have seen that factoring air resistance into projectile motion makes a profound difference upon trajectory when compared to the model that neglects air resistance. For a paintball and a bowling ball launched with sufficient velocity, the range difference can be hundreds of meters. Based on these findings, we can conclude that neglecting air resistance is not a fair assumption. Page 21 of 24
Page 22 of 24 Figure 9: Increasing mass while holding all other variables constant makes the trajectory through air more closely approximate the parabolic trajectory through a vacuum.
Page 23 of 24 Figure 10: Trajectory of a bowling ball launched at 90m/s, 24.
References [1] Arnold, David. Writing Scientific Papers in Latex. College of the Redwoods. http://online.redwoods.edu/instruct/darnold/linalg/latex/project_ latex.pdf [2] Serway, Raymond and Jewett, John. Physics for Scientists and Engineers. Boston, MA: Brooks/Cole, 2014 [3] Pruneau, Claude A. Projectile Motion. Wayne State University. http://rhig.physics.wayne.edu/~pruneau/courses/phy5200/lectures/ PHY5200-Chap2.pdf [4] Kagen, David. Freefall with Air Resistance CSU Chico. http://phys.csuchico. edu/kagan/204a/lecturenotes/section15.pdf [5] Beeken, Paul. First Order Differential Equations Applied to Air Resistance Pretty Good Physics. http://prettygoodphysics.wikispaces.com/file/view/ DifferentialEquations.pdf/273523270/DifferentialEquations.pdf Page 24 of 24