Deriving the Range Equation Ground to Ground Launches
- Slides: 14
Deriving the Range Equation Ground to Ground Launches This works for ground to ground launches only. This derivation uses the principle of symmetry ( what goes up must come down) This derivation is based on the principle that the initial velocity going up will have a negative acceleration until it reaches the paths apex with a velocity of zero The projectile will then accelerate due to gravity until it reaches the ground at a velocity equal to the initial velocity.
Symmetry and Velocities Apex Velocity = 0 m/s Velocity Initial = V 0 Since Velocity Final Equals Velocity Initial then Vf = V 0
Velocity Vector Direction Changes or What Goes Up Must Come Down • In the picture below, the equal in magnitude and opposite in direction velocity vectors are shown as V 0 • The horizontal range is shown as ΔX • The angle of the launch (both assent and descent are shown as θ
Defining the Velocity vector in terms of it’s component parts • If V 0 is treated as the resultant vector, It can be broken into it’s component parts as shown below in blue • Since horizontal velocity is constant then Vx initial = Vx final • There is never an acceleration in the X direction
Defining the Velocity vector in terms of it’s component parts • The second component in the Y direction must be equal and opposite • The first is labeled Voy for Velocity initial y and Vfy for Velocity final y
Defining the Velocity vector in terms of it’s component parts • Since Sinθ = opposite/ hypotenuse then the Opposite side = Hyp * Sinθ • In this case that means Voy = Vo * Sinθ • Since Cosθ = Adjacent/hypotenuse then the Adjacent side = Hyp * Cosθ
Defining the Velocity vector in terms of it’s component parts • Since Horizontal velocity is constant Vx final is the same as Vx initial • In other words Vx final is the same as saying Adjacent side = Hyp * Cosθ • Or Vx = V 0 Cos θ
Defining the Velocity vector in terms of it’s component parts • The Vertical velocity is equal in magnitude but opposite in direction • Vfy is the same as saying the negative of Opposite side = Hyp * Sinθ • Or Vfy = -V 0 Sin θ
Solving the problem The X direction Velocity = Distance / time Vx = Δx/ΔT Since Vx = V 0 Cosθ we can say V 0 Cosθ = Δx/ΔT Solving for time we can say ΔT = Δx/ V 0 Cosθ
Solving the problem The Y direction Typically when speaking of final velocity we have assumed an initial velocity of 0 m/s If that is true Final velocity is simply equal to Acceleration * change in time If our initial velocity is not zero we must add that to this equation and say Vfv = Vo + AΔT In our problem this means Vfv = Voy + AΔT Since Vfv is equal in magnitude but opposite in direction to V oy we can say – Voy = Voy + AΔT Going one step further since Voy = V 0 Sinθ We can substitute – V 0 Sinθ = V 0 Sinθ + AΔT
Combining Terms We now have – V 0 Sinθ = V 0 Sinθ + AΔT Subtract – V 0 Sinθ from both sides we get – 2 V 0 Sinθ = AΔT
Combining Equations From our X direction equation we can substitute for ΔT where ΔT = ΔX/ V 0 Cosθ So – 2 V 0 Sinθ = A(ΔX/ V 0 Cosθ) Now solve for ΔX the range ΔX = – 2 V 20 Sinθ Cosθ A
Trigonometry Identities From Trigonometry we remember 2 Sinθ Cos θ = Sin 2θ Therefore we can substitute and get ΔX = - V 20 Sin 2θ/A Where ΔX = the range
Final Formula The Initial Velocity (v 0) will always be a positive number When the opposite of a positive number is squared Example - V 20 The result is V 20 This leaves a final formula of ΔX = V 20 Sin 2θ/A Where ΔX = the range
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