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Deriving the Range Equation

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/s 2

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Deriving the Range Equation

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  1. Deriving the Range Equation Or, how to get there from here

  2. 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

  3. ymax x x R = 2x

  4. ymax vi q x x R = 2x

  5. vi viy = vi sin q q vx= vicosq

  6. ymax vi viy = vi sin q q x x vx=vicosq R = 2x In the x-direction: , where t = time to top of path t

  7. ymax vi q x x R = 2x In the y-direction:

  8. ymax vi q x x R = 2x In the y-direction: At the top of the path, vfy = 0

  9. ymax vi q x x R = 2x So in the y-direction: Substituting in

  10. ymax vi q x x R = 2x So in the y-direction: Now, we have Substituting in

  11. vi Remember that the initial velocity in the y-direction =vi sin q viy =vi sin q q vx= vicosq

  12. ymax vi q x x R = 2x So we go from To

  13. The whole point here is to solve for x . . . • ( • ( • = x

  14. Remember that the range, R, = 2x = 2x = R Double-angle theorem from trig: so = R (Q.E.D.)

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