1. The range of a projectile is……
how far it goes horizontally
The range depends on the projectile’s…..
the speed and angle fired

2. 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:
The range equation:
2 vi2 cossin = Range
or vi2 sin2 = Range
Solve the vertical equation for t and then substitute it for t into the first equation __ g __ g

3. - (Vi sin θ) ● t = ½ g t2 (now divide both sides by t)
-Vi sin θ = ½ g t (divide ½ g on both sides)
X = (Vi cosθ) ● t = (Vi cosθ) ● (2Vi /g)sin θ
-(2Vi /g)sin θ = t
You can further simplify using the trig identity 2cossin = sin2θ
to get the last form:
Let’s do 2 reality checks:
#1 units
#2 what angle gives the maximum? m 45
The range equation:
2 vi2 cossin = Range
or vi2 sin2 = Range
Solve the vertical equation for t and then substitute it for t into the first equation 2 g __ g

4. Now let’s derive the Trajectory Equation

5. 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:
Now substitute it
t = x / Vicos
2) Subst: y = Visin • ( x / Vicos) - ½ g• ( x / Vicos) 2
Do you see a trig identity that would make this equation less ugly?

6. Which is the general form of ………
y = Vi sin • ( x / Vi cos) - ½ g• ( x / Vicos) 2
y = Vi tan • x - g•x 2 / 2Vi 2 cos 2 
This is of the general mathematical form
y = ax + bx 2
Which is the general form of ………
…..a parabola
y = tan • x - (g/ 2Vi 2 cos 2 ) •x 2

7. Graph this on your graphing calculator
y = tan • x - (g/ 2Vi 2 cos 2 ) •x 2
Calculate the parabolic equation for Vi = 50 m/s and  = 30 degrees
y = tan 30 • x - (10/ 2 (50) 2 cos 2 30 ) •x 2
y = tan 30 • x - (10/ {2 (50) 2 cos 2 30 ) } •x 2
y = • x - 10/3750 •x 2
y = x x 2
Graph this on your graphing calculator
Now recalculate and graph it at 45 degrees
y = tan 45 • x - (10/ {2 (50) 2 cos 2 45 )} •x 2

8. g 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
or vi2 sin2 = R g g
We can derive this from our kinematic using simultaneous equations to eliminate other variables