User:Egm6321.f12.team4.lu/Homework7 R7.4

Problem R7.4 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral with particular values,


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$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (6.10.1)
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$$ a=k/m=2, b=g=10, z(t):=v_y(t) $$
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Initial vertical velocity: $$ z(0)=v_y(0)=50 $$
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n=2 and n=3

For each value of n

(1) The vertical velocity z(t) for different values of time (t) (2) Plot the altitude y(t) vs t (3) Find time where projectile returns to ground. i.e. t when y(t)= 0

Solution
Consider the general equation of a particle with air resistance,


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$$ m.\frac{\mathrm{d} v_y}{\mathrm{d} t}= -k.v_y^{n}- mg $$ (6.10.1)
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where $$ z(t):= v_y(t)$$, there (6.10.1) can also be written as,


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$$ mz^{'}+kz^{n}= -mg $$     (6.10.2)
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Rearranging the terms,


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$$ m.\frac{\mathrm{d} z}{\mathrm{d} t}= -(k.z^n+mg) $$     (6.10.3)
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$$ \frac{m.dz}{k.z^n+mg}= -dt $$     (6.10.4)
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$$ \int \frac{m.dz}{k.z^n+mg}= - \int dt $$ (6.10.5)
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$$ \int \frac{dz}{\frac{k}{m}z^n+g}= - \int dt $$ (6.10.6)
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Case # 1 - The value of n=2, $$ a = k/m = 2 $$ and $$ g=10 $$

(i) To calculate

Solving this equation by hand,


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$$ \int \frac{dz}{2.z^2+10} = - \int dt $$ (6.10.7)
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$$ \frac{1}{10} \int \frac{dz}{\frac{z^2}{5}+1} $$     (6.10.8)
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Consider $$ u =\frac{z}{\sqrt{5}} $$ and hence $$ du= \frac{dz}{\sqrt{5}}$$


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$$ \frac{1}{10} \int \frac{\sqrt {5}du}{u^2+1} $$     (6.10.9)
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$$ \frac{1}{2 \sqrt {5}} \int \frac{1}{u^2+1} $$     (6.10.10)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(u) + constant(k1) $$     (6.10.11)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) + k1 = -t + k2 $$ (6.10.12)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}})-\frac{1}{2 \sqrt {5}} tan^{-1}(\frac{50}{\sqrt{5}}) = -t $$     (6.10.13)
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$$ t= 0.341247 - \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) $$     (6.10.14)
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WolframAlpha Calculation -> http://www.wolframalpha.com/input/?i={1%2F%282*5^0.5%29}{tan+inverse+%2850%2F5^0.5%29}

or to interpret z in terms of t


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$$ z = \sqrt{5}tan(1.52610 - 2\sqrt{5}t) $$     (6.10.15)
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Plot of Velocity, z(t) vs time.

(ii) To find and plot altitude y(t) vs t

The velocity in the y direction, $$ v_y $$ is the first derivative of the altitude y(t), that is $$ v_y=\frac{\mathrm{d} y_t}{\mathrm{d} t}$$


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$$ \frac{\mathrm{d} y_t}{\mathrm{d} t} = \sqrt{5}tan(1.52610- 2\sqrt{5}.t) $$ (6.10.16)
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Integrating with respect to time (t),


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$$ \int_{0}^{y_t}dy_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.17)
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$$ y_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.18)
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WolframAlpha Calculation -> http://www.wolframalpha.com/input/?i=integrate+sqrt%285%29+tan%28%28-10+t%2Bsqrt%285%29+tan^%28-1%29%2810+sqrt%285%29%29%29%2Fsqrt%285%29%29+dt

Plot of Altitude (vertical distance), y(t) vs time.

(iii) To find the time (t) when projectile returns to the ground

As seen in the graph above, t ~ 0.25 (units unknown)

Case # 2 - The value of n=3, $$ a = k/m = 2 $$ and $$ g=10 $$


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$$ \int \frac{dz}{2.z^3+10} = - \int dt $$ (6.10.19)
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we know the initial conditions, as $$ z(t=0)=50 $$


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$$ \int_{50}^{z} \frac{dz}{2.z^3+10} = - \int_{0}^{t} dt $$ (6.10.20)
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Th solution for the above integral can found using

WolframAlpha Calculation -> https://www.wolframalpha.com/input/?i=++integrate+\frac{dz}{2.z^3%2B10}

The solution will also be obtained using a numerical method like the composite rule, more specifically the composite rectangular rule ,

The field of integration is considered to be $$ \left [ a,b \right ]= \left [ 50,z \right ]$$ with the interval as $$ h=\frac{z-50}{2} $$

The two sub intervals, $$ [50,\frac{z-50}{2}]$$ and $$ [\frac{z-50}{2},z]$$

The corresponding midpoints of the two intervals respectively are $$ h_1=\frac{z-150}{2}$$ and $$ h_2=\frac{z+50}{2} $$

The functions as defined at these midpoints are,


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$$ f [\frac{z-150}{2}]= \frac{1}{2(\frac{z-150}{2})^3+10} $$     (6.10.21)
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$$ f [\frac{z+50}{2}]= \frac{1}{2(\frac{z+50}{2})^3+10} $$     (6.10.22)
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Applying the composite rule to calculate the value of the integral
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$$ I_0=h \left ( f(h_1)+f(h_2) \right ) $$     (6.10.23)
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$$ I_0=\frac{z-50}{2}\left [ \frac{1}{2(\frac{z-150}{2})^3+10}+ \frac{1}{2(\frac{z+50}{2})^3+10} \right ] = -t $$     (6.10.24)
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$$ t=\frac{50-z}{2}\left [ \frac{1}{2(\frac{z-150}{2})^3+10}+ \frac{1}{2(\frac{z+50}{2})^3+10} \right ] $$     (6.10.25)
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which is the value of t and velocity z(t)

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by --