1.3 The range and trajectory equations DERIVED

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The range of a projectile is……
how far it goes horizontally
The range depends on the projectile’s….. the speed and angle fired
The range equation:
2 vi2 cossin
g
or vi2 sin2 = R
g
We can derive this from
our kinematic using
simultaneous equations to
eliminate other variables
Analyzing the x and y motion gives us 2 simultaneous equations
X = Xo + (Vi cosθ) ● t
(there is no acceleration horizontally)
Y = Yo + (Vi sin θ) ● t + ½ g t2 (g is 9.8 m/s2)
Now lets eliminate variables until we get the equations below
1. Get rid of the Xo and Yo: set your reference frame to start at the origin
2. Get rid Y:
You land at Yo again, which equals zero height
3. Get rid of t:
Solve the vertical equation for t and
then substitute it for t into the first
equation
The range equation:
2__vi2 cossin = Range
g
or vi2 sin2 = Range
__
g
- (Vi sin θ) ● t = ½ g t2 (now divide both sides by t)
-Vi sin θ = ½ g t
-(2Vi /g)sin θ = t
(divide ½ g on both sides)
X = (Vi cosθ) ● t = (Vi cosθ) ● (2Vi /g)sin θ
You can further simplify using the trig identity cossin = sin2θ
to get the last form:
Let’s do 2 reality checks: m
#1 units
45
#2 what angle gives the maximum?
Solve the vertical equation for t and
then substitute it for t into the first
equation
The range equation:
2__vi2 cossin = Range
g
or vi2 sin2 = Range
__
g
Now let’s derive
the Trajectory
Equation
We have shown, for projectiles;
x(t) = Vicos· t and y(t) = Visin• t - ½ g•t2
How can we write y= f(x)?
How can we remove time from the equations?
Creating a trajectory equation
1) Eliminate t. Solve the first equation for t and substitute it wherever t
appears in the second equation.
Solve for t:
t = x / Vicos
Now substitute it
2) Subst: y = Visin • ( x / Vicos) - ½ g• ( x / Vicos)
2
Do you see a trig identity that would make this equation less ugly?
y = Vi sin • ( x / Vi cos) - ½ g• ( x / Vicos)
y = Vi tan • x - g•x
2
2
/ 2Vi 2 cos 2 
This is of the general mathematical form
y = ax + bx 2
Which is the general form of ………
y = tan • x - (g/ 2Vi 2 cos 2 ) •x
2
…..a parabola
y = tan • x - (g/ 2Vi 2 cos 2 ) •x
2
Calculate the parabolic equation for Vi = 50 m/s and  = 30, 45 and 60 degrees
y = tan 30 • x - (10/ 2 (50) 2 cos 2 30 ) •x
y = tan 30 • x - (10/ 2 (50) 2 cos
2
2
30 ) •x 2
0.577• x - (0.75) •x 2
0.577 x - .75x 2
Graph this on your graphing calculator
y =
y =
Now recalculate and graph it at 45 degrees
y = tan 45 • x - (10/ 2 (50) 2 cos 2 45 ) •x
Now recalculate and graph it at 60 degrees
2

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