Projectile Motion
This blog is about projectile motion. Most resources give an introductory treatment of projectile motion. And yes, I will present the non-air resistance model in this section. But I am also going to present a more realistic model, taking into account air resistance, size of the projectile, and temperature.
All amounts are presented in SI (kilograms, meters, seconds) units.
Quick Conversions
1 ft = 3.280839895 m
1 lb = 0.4535924 kg
Simple Projectile Motion
A perfect projectile can be described by the following equations:
x(t) = v0 t cos θ
y(t) = v0 t sin θ - 0.5 g t^2
Where
v0 = initial velocity
θ = Intial angle, in degrees
g = gravitational acceleration constant = 9.80665 m/s^2 = 32.17404856 ft/s^2
And we assume the starting height is zero (on the ground)
Note that air resistence, features of the object, and other outside objects are ignored.
Example: Hitting a Golf Ball
Use the simplistic model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle.
The trajectory lasts for approximately 7.844 seconds, with a range of approximately 430.86 meters and a height of approximately 75.4 meters.
A More Realisitc View of Projectile Motion
The equations above describe simple projectile motion. However, if we want a realistic picture, we have to consider factors such as temperature, air pressure, and the size of the projectile.
So of the variables we will consider are the drag coefficient, the density of air, and the object's terminal velocity.
Note: Linear drag is assumed. This is best for objects moving at slow speeds. No lift is used.
Drag Coefficient
The drag coefficient is a dimensionless quantity that quantifies the resistance of an objects's movement, taking into consideration factors such as the object's size, shape, and smoothness.
Approximate Drag Coefficients:
Perfect Smooth Baseball: 0.1
Baseball: 0.3
Golf Ball: 0.4
Tennis Ball: 0.6
Density of Air
The density of air can be determined using the Ideal Gas Law:
P V = N R T
where
P = absolute pressure = 101325 kPa
R = Specific Gas Constant = 287.058 J/kg K
T = temperature in Kelvins
N = number of moles (mass)
V = volume of the gas
ρ = air density = N/V in kg/m^3
Solving for ρ gives:
ρ = P / (R T) ≈ 352.9774471/T
Note: To convert degrees Celsius to Kevlins, add 273.15.
Terminal Velocity
When an object falls, it reaches a terminal velocity when the drag becomes equal to the object's weight. At terminal velocity, the object no longer accelerates.
Terminal Velocity is measured by:
V = √ ( ( 2 × mass of object × g ) / ( drag coefficient × ρ × surface area of object ) )
TROJECT
Calculator: CASIO fx-CG10 (Prizm)
Note: Let [triangle] be the symbol for the "stop and display triangle"
{.3, .6, .4} → List "C"
{.145, .057, .045} → List "M"
{.041043, .003318, .001385} → List "S"
Deg
ClrText
"Init Velocity="?→ V
"Init Angle="? → θ
9.80665 → G
Menu "Object","Baseball",1,"Tennis Ball",2,"Golf Ball",3
Lbl 1: 1 → I : Goto Y
Lbl 2 : 2 → I : Goto Y
Lbl 3 : 3 → I : Goto Y
Lbl Y
Menu "Temperature","104°F/40°C",A,"95°F/35°C",B,
"86°F/30°C",C,"77°F/25°C",D,
"68°F/20°C",E,"59°F/15°C",F,
"50°F/10°C",G,"41°F/5°C",H,
"32°F/0°C",I
Lbl A : 1.1839 → P : Goto Z
Lbl B : 1.2041 → P : Goto Z
Lbl C : 1.2250 → P : Goto Z
Lbl D : 1.2466 → P : Goto Z
Lbl E : 1.2690 → P : Goto Z
Lbl F : 1.2920 → P : Goto Z
Lbl G : 1.3163 → P : Goto Z
Lbl H : 1.3413 → P : Goto Z
Lbl I : 1.3943 → P : Goto Z
Lbl Z
List "M"[I] → M
√ (( 2 × M × G) ÷ (List "C"[I] × P × List "S"[I])) → W
M × G ÷ W → K
Solve(-MGX ÷ K + M ÷ K × ( 1 - e^(-KX ÷ M)),100) → E
MV cos θ ÷ K × ( 1 - e^(-KE ÷ M)) → R
"Range="
R [triangle]
-M ÷ K × ln ((MG) ÷ K × (V sin θ + MG ÷ K)^-1) → F
-MGF ÷ K + M ÷ K × (V sin θ + MG ÷ K) × (1 - e^(-KF ÷ M)) → H
"Height="
H [triangle]
ClrGraph
ViewWindow -.5, R + .5, 1, -1, H+1, 1, 0, E, .05
θ → A
ParamType
G SelOff
G SelOn 1
"MV cos A ÷ K × (1 - e^(-KT ÷ M))" → Xt1
"-MGT ÷ K + M ÷ K × (V sin A + MG ÷ K) × (1 - e^(-KT ÷ M))" → Yt1
DrawGraph
-----------
Variables:
M = mass of object, in grams
V = initial velocity, in metes/seconds
A = θ = initial angle, in degrees
C = drag coefficient
S = radius of the object
P = gas density
K = terminal velocity
Revisiting the Example: The Golf Ball
Use the proposed model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle. Assume it is 77°F outside (25°C).
The trajectory has a range of approximately 163.46 meters and a height of approximately 45.02 meters. It lasts about 6.177 seconds.
References:
Article: Erlichson, Herman. "Maximum Projectile range with drag and lift, with particular application to golf." The College of Staten Island, 1982
From:
Armenti, Angelo Jr. (editor) The Physics of Sports American Institute of Physics 1983 pg. 71-78
Reference Links:
Terminal Velocity (NASA.gov)
Drag Coefficient (NASA.gov)
Drag on a Baseball (NASA.gov)
Air Pressure
