Trajectory of A Projectile - Trajectory of A Projectile With Air Resistance

Trajectory of A Projectile With Air Resistance

Air resistance will be taken to be in direct proportion to the velocity of the particle (i.e. ). This is valid at low speed (low Reynolds number), and this is done so that the equations describing the particle's motion are easily solved. At higher speed (high Reynolds number) the force of air resistance is proportional to the square of the particle's velocity (see drag equation). Here, and will be used to denote the initial velocity, the velocity along the direction of x and the velocity along the direction of y, respectively. The mass of the projectile will be denoted by m. For the derivation only the case where is considered. Again, the projectile is fired from the origin (0,0).

For this assumption, that air resistance may be taken to be in direct proportion to the velocity of the particle is not correct for a typical projectile in air with a velocity above a few tens of meters/second, and so this equation should not be applied to that situation.

The free body diagram on the right is for a projectile that experiences air resistance and the effects of gravity. Here, air resistance is assumed to be in the direction opposite of the projectile's velocity. (actually is more realistic, but not used here, to ensure an analytic solution,) is written due to the initial assumption of direct proportionality implies that the air resistance and the velocity differ only by a constant arbitrary factor with units of N*s/m.

As an example, say that when the velocity of the projectile is 4 m/s, the air resistance is 7 newtons (N). When the velocity is doubled to 8 m/s, the air resistance doubles to 14 N accordingly. In this case, k = 7/4 N x s/m. Note that k is needed in order to relate the air resistance and the velocity by an equal sign: otherwise, it would be stating incorrectly that the two are always equal in value (i.e. 1 m/s of velocity gives 1 N of force, 2 m/s gives 2 N etc.) which isn't always the case, and also it keeps the equation dimensionally correct (a force and a velocity cannot be equal to each other, e.g. m/s = N). As another quick example, Hooke's Law describes the force produced by a spring when stretched a distance x from its resting position, and is another example of a direct proportion: k in this case has units N/m (in metric).

To show why k = 7/4 N·s/m above, first equate 4 m/s and 7 N:

(Incorrect)

(Introduction of k)

( cancels)

For more on proportionality, see: Proportionality (mathematics)

The relationships that represent the motion of the particle are derived by Newton's Second Law, both in the x and y directions. In the x direction and in the y direction .

This implies that: (1),

and

(2)
Solving (1) is an elementary differential equation, thus the steps leading to a unique solution for and, subsequently, will not be enumerated. Given the initial conditions (where is understood to be the x component of the initial velocity) and for :

(1a)

(1b)

While (1) is solved much in the same way, (2) is of distinct interest because of its non-homogeneous nature. Hence, we will be extensively solving (2). Note that in this case the initial conditions are used and when .

(2)

(2a)

This first order, linear, non-homogeneous differential equation may be solved a number of ways, however, in this instance it will be quicker to approach the solution via an integrating factor: .

(2c)

(2d)

(2e)

(2f)

(2g)

And by integration we find:

(3)

Solving for our initial conditions:

(2h)

(3a)

With a bit of algebra to simplify (3a):
(3b)

An example is given using values for the mass and terminal velocity for a baseball taken from .

m = 0.145 kg (5.1 oz)

v₀ = 44.7 m/s (100 mph)

g = -9.81 m/s² (-32.2 ft/s²)

v_t = -33.0 m/s (-73.8 mph)

.

The red path (the lower path) is the path taken by the projectile modeled by the equations derived above, and the green path is taken by an idealized projectile, one that ignores air resistance altogether. (3.28 ft/m) Ignoring air resistance is not ideal in this scenario, as with no air resistance, a home run could be hit with 270 ft to spare. (The mechanics of pitching at 45 degrees notwithstanding.) And in most cases it's more accurate to assume, meaning when air resistance increases by a factor of p the resistance increases by . In the first example of proportionality, where the velocity was doubled to 8 m/s, the air resistance would instead be quadrupled to 28 N: this only adds to the large amount of error in neglecting air resistance. For an analytic solution: Shouryya Ray solves one of Newton's puzzles