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 v i 2 cos  sin  g or

v

i 2

sin2



g = R

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 t 2 (g is 9.8 m/s 2 ) 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 v i 2 cos  sin  = Range g or

v

g i 2

sin2



= Range

- (Vi sin θ) ● t = ½ g t 2 (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 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 v i 2 cos  sin  = Range g or

v

g i 2

sin2



= Range

Now let’s derive the Trajectory Equation

We have shown, for projectiles; x(t) = V i cos  · t and y(t) = V i sin  • t - ½ g•t 2 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 / V i cos  Now substitute it 2) Subst: y = V i sin  • ( x / V i cos  ) - ½ g• ( x / V i cos  ) 2 Do you see a trig identity that would make this equation less ugly?

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

y = tan  • x (g/ 2V i 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 2 y = tan 30 • x (10/ 2 (50) 2 cos 2 30 ) •x 2 y = 0.577

• y = 0.577

x ( 0.75) •x 2 x - .75x 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 Now recalculate and graph it at 60 degrees