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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 cossin 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 cossin = 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 cossin = 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 cossin = 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