theEric
  • theEric
(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.
Physics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

theEric
  • 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
  • anonymous
I was playing with trig identities to try to get the angle by itself
theEric
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

theEric
  • 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
  • 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
  • anonymous
yes that path should work, going to be a messy equation with the trig functions though
theEric
  • theEric
Thanks! I'm typing my work out as I go, and I'll post it so anyone can see.
theEric
  • 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
  • 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
  • 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
  • anonymous
well you have the correct setup here, there maybe a trig identity somewhere for this
anonymous
  • anonymous
I found this identify: \[sin(u)cos(v) = \frac{sin(u+v) +sin(u-v)}{2}\]
theEric
  • theEric
Thanks! I don't see where to use it yet, though.
anonymous
  • anonymous
I used it before taking the derivative with respect to theta
theEric
  • theEric
I see, thanks!
anonymous
  • anonymous
I get \[\theta = 1.231 radians\]
anonymous
  • anonymous
Were you able to solve?
theEric
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
this is where \[\theta = 1.231rad\]
theEric
  • 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
  • theEric
Wait,\[3\cos\left(\frac{3}{2}\theta\right) = \cos\left(\frac{-1}{2}\theta\right)\]
anonymous
  • anonymous
missing a negative sign you get a double negative on the right side
theEric
  • 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
  • anonymous
that is what I got and that is the 1.231rad answer
theEric
  • theEric
Thanks!
theEric
  • theEric
I think I'll stop there for tonight! Thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.