- #1
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):
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_{x} = \frac{dv_{x}}{dt}
and
-.25v_{y} -g = \frac{dv_{y}}{dt}
Therefore
\frac{dv_{y}}{dt} +.25 v_{y} = -g where g is the Earth's acceleration due to gravity.
and
\frac{dv_{x}}{dt} +.25v_{x} = 0
Integrating factor: e^{\int P(t) dt}
x range = v_{i}cosθ * t
For v_{x}:
I(t) = e^{\int P(t) dt}
I(t) = e^{.25t + k_{1}}
\int(d e^{.25t}e^{k_{1}}v_{x} /dt) = \int 0 dt
e^{.25t}e^{k_{1}}v_{x} = C_{1}
v_{x} = C_{1}e^{.25t} because e^{k_{1}} is just a constant too.
v_{i}cos(\Theta) = C_{1}e^{.25t}
I use the statutory initial velocity of a golf ball of 76.2 m/s.
cos(\Theta) = \frac{C_{1}}{76.2}e^{.25t}
\Theta = cos^{-1}(\frac{C_{1}}{76.2}e^{.25t})
For v_{y}: (skipping the prelim stuff)
e^{.25t}e^{k_{2}}v_{y} = -gt + C_{2}
v_{y} = e^{-.25t}e^{-k_{2}}(-gt + C_{2})
v_{i}sin(θ) = e^{-.25t}e^{-k_{2}}(-gt + C_{2})
sin(θ) = \frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}
θ = sin^{-1}(\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2})
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_{y} differential equation? Laplace Transform?
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_{x} = \frac{dv_{x}}{dt}
and
-.25v_{y} -g = \frac{dv_{y}}{dt}
Therefore
\frac{dv_{y}}{dt} +.25 v_{y} = -g where g is the Earth's acceleration due to gravity.
and
\frac{dv_{x}}{dt} +.25v_{x} = 0
Homework Equations
Integrating factor: e^{\int P(t) dt}
x range = v_{i}cosθ * t
The Attempt at a Solution
For v_{x}:
I(t) = e^{\int P(t) dt}
I(t) = e^{.25t + k_{1}}
\int(d e^{.25t}e^{k_{1}}v_{x} /dt) = \int 0 dt
e^{.25t}e^{k_{1}}v_{x} = C_{1}
v_{x} = C_{1}e^{.25t} because e^{k_{1}} is just a constant too.
v_{i}cos(\Theta) = C_{1}e^{.25t}
I use the statutory initial velocity of a golf ball of 76.2 m/s.
cos(\Theta) = \frac{C_{1}}{76.2}e^{.25t}
\Theta = cos^{-1}(\frac{C_{1}}{76.2}e^{.25t})
For v_{y}: (skipping the prelim stuff)
e^{.25t}e^{k_{2}}v_{y} = -gt + C_{2}
v_{y} = e^{-.25t}e^{-k_{2}}(-gt + C_{2})
v_{i}sin(θ) = e^{-.25t}e^{-k_{2}}(-gt + C_{2})
sin(θ) = \frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}
θ = sin^{-1}(\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2})
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_{y} differential equation? Laplace Transform?
