Deriving the Range Equation Ground to Ground Launches

Download Report

Transcript Deriving the Range Equation Ground to Ground Launches

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 = 0m/s
Velocity
Initial = V0
Since Velocity Final
Equals Velocity
Initial then Vf = V0
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 V0
• 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 V0 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 = V0 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 = -V0 Sin θ
Solving the problem
The X direction






Velocity = Distance / time
Vx = Δx/ΔT
Since Vx = V0 Cosθ we can say
V0 Cosθ = Δx/ΔT
Solving for time we can say
ΔT = Δx/ V0 Cosθ
Solving the problem
The Y direction
 Typically when speaking of final velocity we have assumed an
initial velocity of 0m/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 Voy we
can say – Voy = Voy + AΔT
 Going one step further since Voy = V0 Sinθ
 We can substitute – V0 Sinθ = V0 Sinθ + AΔT
Combining Terms
 We now have – V0 Sinθ = V0 Sinθ + AΔT
 Subtract – V0 Sinθ from both sides we get
 –2V0 Sinθ = AΔT
Combining Equations
 From our X direction equation we can substitute for
ΔT where ΔT = ΔX/ V0 Cosθ
 So –2V0 Sinθ = A(ΔX/ V0 Cosθ)
 Now solve for ΔX the range
 ΔX = –2V20 Sinθ Cosθ
A
Trigonometry Identities
 From Trigonometry we remember
 2 Sinθ Cos θ = Sin 2θ
 Therefore we can substitute and get
 ΔX = - V20 Sin 2θ/A
 Where ΔX = the range
Final Formula
 The Initial Velocity (v0) will always be a positive
number
 When the opposite of a positive number is squared




Example - V20
The result is V20
This leaves a final formula of ΔX = V20 Sin 2θ/A
Where ΔX = the range