The range of a projectile is……

The range depends on the projectile’s…..

how far it goes horizontally

the speed and angle fired

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)

The range equation:

2 vi2 cossin = Range

or vi2 sin2 = Range

__ g

__ g

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 cossin = Range

or vi2 sin2 = Range

2 g

__ g

Solve the vertical equation for t and then substitute it for t into the first equation

- (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)

-(2Vi /g)sin θ = t X = (Vi cosθ) ● t = (Vi cosθ) ● (2Vi /g)sin θ

You can further simplify using the trig identity 2cossin = sin2θ to get the last form:

Let’s do 2 reality checks: #1 units #2 what angle gives the maximum?

m45

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?

2) Subst: y = Visin • ( x / Vicos) - ½ g• ( x / Vicos) 2

Do you see a trig identity that would make this equation less ugly?

Creating a trajectory equation

Solve for t: t = x / Vicos

1) Eliminate t. Solve the first equation for t and substitute it wherever t appears in the second equation.

Now substitute it

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

y = tan • x - (g/ 2Vi 2

cos 2 ) •x 2

Calculate the parabolic equation for Vi = 50 m/s and = 30 degrees

y = 0.577• x - 10/3750 •x 2

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 = 0.577 x - 0.00267x 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

The range of a projectile is……

The range depends on the projectile’s…..

The range equation:

2 vi2 cossin

or vi2 sin2 = R

g

g

how far it goes horizontally

the speed and angle fired

We can derive this from our kinematic using simultaneous equations to eliminate other variables