Golf Balls and differential equations

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SUMMARY

This discussion focuses on modeling the trajectory of a golf ball using differential equations, specifically addressing the equations for velocity without considering lift. The drag coefficient is set at -0.25, leading to the equations: -0.25v_{x} = dv_{x}/dt and -0.25v_{y} - g = dv_{y}/dt. The user attempts to derive the x-range as a function of the launch angle θ, utilizing integration techniques and initial conditions based on a standard golf ball velocity of 76.2 m/s. The challenge lies in expressing the initial velocity in the y-direction as a function of θ and determining the constants involved.

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  • Knowledge of projectile motion principles, including the effects of gravity and drag.
  • Basic proficiency in using mathematical notation and functions, such as trigonometric functions.
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  • Explore the application of Laplace Transforms in solving differential equations related to projectile motion.
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Hercuflea
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Hey Everybody, I am supposed to model the trajectory of a golf ball. I have been given the equations for velocity as a function of its derivative with respect to time. I am supposed to find the x-range as a function of the angle θ. (Pardon my bad latex skills, I will fix mistakes):

Homework Statement


These are the equations which have a mathematical solution, and they do not include lift. -.25 is the drag coefficient on the golf ball.


-.25v[itex]_{x}[/itex] = [itex]\frac{dv_{x}}{dt}[/itex]
and
-.25v[itex]_{y}[/itex] -g = [itex]\frac{dv_{y}}{dt}[/itex]

Therefore

[itex]\frac{dv_{y}}{dt}[/itex] +.25 v[itex]_{y}[/itex] = -g where g is the Earth's acceleration due to gravity.
and
[itex]\frac{dv_{x}}{dt}[/itex] +.25v[itex]_{x}[/itex] = 0

Homework Equations



Integrating factor: e[itex]^{\int P(t) dt}[/itex]
x range = v[itex]_{i}[/itex]cosθ * t

The Attempt at a Solution



For v[itex]_{x}[/itex]:

I(t) = e[itex]^{\int P(t) dt}[/itex]
I(t) = e[itex]^{.25t + k_{1}}[/itex]

[itex]\int(d e^{.25t}e^{k_{1}}v_{x} /dt)[/itex] = [itex]\int 0 dt[/itex]

e[itex]^{.25t}[/itex]e[itex]^{k_{1}}[/itex]v[itex]_{x}[/itex] = C[itex]_{1}[/itex]

v[itex]_{x}[/itex] = C[itex]_{1}[/itex]e[itex]^{.25t}[/itex] because e[itex]^{k_{1}}[/itex] is just a constant too.

v[itex]_{i}[/itex]cos([itex]\Theta[/itex]) = C[itex]_{1}[/itex]e[itex]^{.25t}[/itex]

I use the statutory initial velocity of a golf ball of 76.2 m/s.

cos([itex]\Theta[/itex]) = [itex]\frac{C_{1}}{76.2}[/itex]e[itex]^{.25t}[/itex]

[itex]\Theta[/itex] = cos[itex]^{-1}[/itex]([itex]\frac{C_{1}}{76.2}[/itex]e[itex]^{.25t}[/itex])

For v[itex]_{y}[/itex]: (skipping the prelim stuff)

e[itex]^{.25t}[/itex]e[itex]^{k_{2}}[/itex]v[itex]_{y}[/itex] = -gt + C[itex]_{2}[/itex]

v[itex]_{y}[/itex] = e[itex]^{-.25t}[/itex]e[itex]^{-k_{2}}[/itex](-gt + C[itex]_{2}[/itex])

v[itex]_{i}[/itex]sin(θ) = e[itex]^{-.25t}[/itex]e[itex]^{-k_{2}}[/itex](-gt + C[itex]_{2}[/itex])

sin(θ) = [itex]\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}[/itex]

θ = sin[itex]^{-1}[/itex]([itex]\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}[/itex])

These equations for θ seem pretty nasty, not to mention I have no way of knowing the Constants because I only know the absolute value of the velocity, not the components.
Also, these equations I have found for θ have seemingly nothing to do with range, they are a function of time. Any hints? Should I use another solution method for the v[itex]_{y}[/itex] differential equation? Laplace Transform?
 
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I tried to solve v[itex]_{y}[/itex] using Laplace Transforms.

I got

v[itex]_{y}[/itex] = e[itex]^{-.25t}[/itex](v[itex]_{y}[/itex](0) +4g) - 4g

I still don't know v[itex]_{y}[/itex](0) because the angle can vary.
 
Getting closer, can someone tell me if there is a way to write initial velocity in y as a function of θ?
 

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