(Classical Mechanics (with calc, but algebra could help)) Hi! I am stumped, but I feel like I should be able to get to the answer... A pitcher has a throwing speed that is a function of the angle the ball is thrown at. The problem is below, along with what I've thought of.

- theEric

- schrodinger

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- theEric

\(\sf\Large\color{orange}{The\ Problem}\)
"A baseball pitcher can throw a ball more easily horizontally than vertically. Assume that the pitcher's throwing speed varies with elevation angle approximately as \(v_0cos\dfrac{1}{2}\theta_0\), where \(\theta_0\) is the initial elevation angle and \(v_0\) is the initial velocity when the ball is thrown horizontally.
Find the angle \(\theta_0\) at which the ball must be thrown to achieve maximum (problem a:) height and (problem b) range."
That's the part I need help with..
\(\sf\Large\color{orange}{My\ thoughts...}\)
I've thought of using \(v_y = \left(v_0\ \cos\left(\dfrac{1}{2} \theta_0\right)\right)\sin\theta_0+gt\) and setting \(v_y=0\). I tried to do the same with \(v_y^2=v_0\ \cos\left(\dfrac{1}{2}\theta_0\right)+2gh\), but to no avail. I also thought of looking at \(y\) and finding the maximum of its function, but that looks to dificult.
I'm a slow learner, but I like to learn! Thanks for any help!

- anonymous

I was playing with trig identities to try to get the angle by itself

- theEric

I have to solve for what \(\theta_0\) would be to attain a maximum height. And, after that, I have to solve for a \(\theta_0\) that would give the maximum range. They will both be functions of \(v_0\). I should be able to calculate \(\theta_0\) in each case by knowing \(v_0\) and using substitution.

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## More answers

- theEric

That's a good idea! I'll look at that. It was a fleeting thought earlier. I'll look for anything that can help...

- theEric

I actually got to a point a while ago when I was looking at a function for \(y\) and finding its maximum. I took the derivative withe respect to \(\theta\) (which went in the place of \(\theta_0\)) and set it to 0 to find extrema. I was going to look to see which extrama values for \(\theta\) would be within my acceptable range of \(\theta_0\): from 0 to 90 degrees (the way a pitcher would throw).
And I found an identity to help! :)

- anonymous

yes that path should work, going to be a messy equation with the trig functions though

- theEric

Thanks! I'm typing my work out as I go, and I'll post it so anyone can see.

- theEric

Letting \(g\approx -9.8\ [\text m/\text s]\)
\(y=\dfrac{1}{2}gt^2+v_0t\cos\left(\dfrac{1}{2}\theta_0\right)\sin\theta_0\)
So \(y\) is at a maximum when \(\cos\left(\dfrac{1}{2}\theta_0\right)\sin\theta_0\) is a maximum. So I let \(\theta\) be variable and be in the place of \(\theta_0\) so I can find \(\theta_0\).
The extrema of \(y\) (maximum included) are at the \(\theta\) where \(\dfrac{d}{d\theta}\left(\cos\left(\dfrac{1}{2}\theta\right)\sin\theta\right)=0\).
\[\dfrac{d}{d\theta}\left(\cos\left(\dfrac{1}{2}\theta\right)\sin\theta\right)
=\dfrac{-1}{2}\sin\left(\dfrac{1}{2}\theta\right)\sin\left(\theta\right)+\cos\left(\dfrac{1}{2}\theta\right)\cos\left(\theta\right)
=0\]
Awww... Not quite.
\[\cos(u\pm v)=\cos(u)\cos(v)\pm\sin(u)\sin(v)\\
\longrightarrow
\cos(u + v)=\cos(u)\cos(v)+\sin(u)\sin(v)\]
Which is close with a \(u=\dfrac{1}{2}\theta\) and \(v=\theta\).

- theEric

Unfortunately, what I have is more like
\[\cos(u\pm v)\neq\cos(u)\cos(v)\pm\dfrac{-1}{2}\sin(u)\sin(v)\\\]

- theEric

I will have to think of a different, method, I think. These problems are from my book. They tend to require the student to have some skill, and use knowledge flexibly. But they're usually not so in-depth as to make the student find extrema as I tried. I don't think that is what the book would have me do.

- anonymous

well you have the correct setup here, there maybe a trig identity somewhere for this

- anonymous

I found this identify:
\[sin(u)cos(v) = \frac{sin(u+v) +sin(u-v)}{2}\]

- theEric

Thanks! I don't see where to use it yet, though.

- anonymous

I used it before taking the derivative with respect to theta

- theEric

I see, thanks!

- anonymous

I get \[\theta = 1.231 radians\]

- anonymous

Were you able to solve?

- theEric

\(y=\dfrac{1}{2}gt^2+v_0t\cos\left(\dfrac{1}{2}\theta\right)\sin\theta\\\\\large
=\dfrac{1}{2}gt^2+v_0t\dfrac{\sin(\theta+\frac{1}{2}\theta) +\sin(\theta-\frac{1}{2}\theta)}{2}\)
Now I can focus on \(\large \sin(\theta+\frac{1}{2}\theta) +\sin(\theta-\frac{1}{2}\theta)\)
\(\dfrac{d}{d\theta}\left(\large \sin(\theta+\frac{1}{2}\theta) +\sin(\theta-\frac{1}{2}\theta)\right)\\
=\dfrac{d}{d\theta}\left(\large \sin(\theta\left(1+\frac{1}{2}\right)) +\sin(\theta\left(1-\frac{1}{2}\right))\right)\\
=\left(\frac{3}{2}\right) \sin(\theta\left(\frac{3}{2}\right))
+
\left(\frac{-1}{2}\right) \sin(\theta\left(\frac{-1}{2}\right))
\)
...
Cool, thanks! Hopefully I can arrive at that! It does satisfy my interval, \(\theta_0\in \left(0,\ \dfrac{\pi}{2}\right)\).

- anonymous

I was not able to solve for the angle alone on one side of the equation, instead I had to plot two functions and find where they equal

- theEric

Thanks! My teacher will probably want me to solve algebraically, so I'll see if I can do that. I still imagine there must be a quicker way.

- anonymous

when you start using trig functions they always make things messy, you will not get credit if you use calculus?? that seems silly

- theEric

It does get messy. Calculus is fine, too!
I used Wolfram Alpha for \[3sin\left(\frac{3}{2}\theta\right)=sin\left(\frac{-1}{2}\theta\right)\]and there was no \(\theta\) in range, for me...
http://www.wolframalpha.com/input/?i=3sin%283%2F2+*+theta%29%3Dsin%28-1%2F2+*+theta%29

- anonymous

I took the derivative still after using the trig identify and found
\[3cos(\frac{3}{2}\theta) = -cos(\frac{1}{2}\theta)\]
http://www.wolframalpha.com/input/?i=3cos%283%2F2+*+theta%29%3D-cos%281%2F2+*+theta%29

- anonymous

this is where \[\theta = 1.231rad\]

- theEric

Oh! I made a mistake, and corrected it to get
\[3\cos\left(\frac{3}{2}\theta\right) = \cos\left(\frac{1}{2}\theta\right) \]

- theEric

Wait,\[3\cos\left(\frac{3}{2}\theta\right) = \cos\left(\frac{-1}{2}\theta\right)\]

- anonymous

missing a negative sign you get a double negative on the right side

- theEric

\[\dfrac{d}{d\theta}\left(\large \sin(\theta+\frac{1}{2}\theta) +\sin(\theta-\frac{1}{2}\theta)\right)\]\[\large=\dfrac{d}{d\theta}\left[ \sin\left(\theta(1+\frac{1}{2})\right) +\sin\left(\theta(1-\frac{1}{2})\right)\right]\]\[\large=\dfrac{d}{d\theta}\left[ \sin\left(\frac{3}{2}\theta\right) +\sin\left(\frac{1}{2}\theta\right)\right]\]
\[\large=\frac{3}{2}\cos\left(\frac{3}{2}\theta\right) +\frac{1}{2} \cos\left(\frac{1}{2}\theta\right)\]\[\large=\frac{1}{2}\left[3\cos\left(\frac{3}{2}\theta\right) + \cos\left(\frac{1}{2}\theta\right)\right] =0\]
\[\large\implies 3\cos\left(\frac{3}{2}\theta\right) + \cos\left(\frac{1}{2}\theta\right)=0\]\[\large\implies 3\cos\left(\frac{3}{2}\theta\right) =- \cos\left(\frac{1}{2}\theta\right)\]
That is \(\sf\color{blue}{correction~number~2}\)... So I got a negative and dropped a negative sign and gained a negative sign.

- anonymous

that is what I got and that is the 1.231rad answer

- theEric

Thanks!

- theEric

I think I'll stop there for tonight! Thank you!

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