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The range of a projectile is……

The range of a projectile is……. h ow 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  or v i 2 sin2 = R. g. g.

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The range of a projectile is……

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  1. 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 vi2cossin or vi2sin2 = R g g We can derive this from our kinematic using simultaneous equations to eliminate other variables

  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 vi2cossin = Range or vi2sin2 = 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 cos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 vi2cossin = Range or vi2sin2 = Range Solve the vertical equation for t and then substitute it for t into the first equation __ 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. y = Vi sin•( x /Vi cos) - ½ g• ( x /Vicos)2 y = Vi tan•x - g•x 2 /2Vi 2cos 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 2cos 2)•x 2

  7. y = tan •x - (g/2Vi 2cos 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)2cos 230 )•x 2 y = tan 30•x - (10/2 (50)2cos 230 )•x 2 y = 0.577•x - (0.75)•x 2 y = 0.577x - .75x 2 Graph this on your graphing calculator Now recalculate and graph it at 45 degrees y = tan 45•x - (10/2 (50)2cos 245 )•x 2 Now recalculate and graph it at 60 degrees

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