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Deriving the Range Equation Or, how to get there from here Keep in mind . . . • Horizontal velocity REMAINS CONSTANT – No net force is acting horizontally so there is no horizontal acceleration • Vertical velocity CHANGES – Acceleration due to gravity, ~9.81 m/s2 – Caused by the unbalanced force of gravity acting on the object ymax x x R = 2x ymax vi q x x R = 2x vi viy = vi sin q q vx = vi cos q ymax vi q viy = vi sin q x vx =vi cos q x R = 2x In the x-direction: = , where t = time to top of path = ( cos )t = cos ymax vi q x x R = 2x In the y-direction: = + , ℎ = 9.81 / 2 ymax vi q x x R = 2x In the y-direction: = + , ℎ = 9.81 / 2 At the top of the path, vfy = 0 ymax vi q x x R = 2x So in the y-direction: 0 = + Substituting in = cos ymax vi q x x R = 2x So in the y-direction: 0 = + Substituting in = cos Now, we have 0 = + cos Remember that the initial velocity in the y-direction = vi sin q vi viy = vi sin q q vx = vi cos q ymax vi q x x R = 2x So we go from 0 = + cos To 0 = sin + cos The whole point here is to solve for x . . . • − sin = cos • (− sin ) cos = 2sin cos ) • (− • − 2 sin cos =x ( cos = cos ) Remember that the range, R, = 2x −2 2 sin cos = 2x = R Double-angle theorem from trig: 2 sin cos = sin 2 so −2 sin 2 =R (Q.E.D.)