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theEric
 one year ago
(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
 one year ago
(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.

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theEric
 one year ago
Best ResponseYou've already chosen the best response.0\(\sf\Large\color{orange}{The\ Problem}\) "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 \(y(t)\) and finding the maximum of its function, but that looks to difficult. I'm a slow learner, but I like to learn! Thanks for any help!

PhysicsGuru
 one year ago
Best ResponseYou've already chosen the best response.1(A) You can use \[V_{yfinal}^2 = V_{yinitial}^2 + 2(g)(Y_{final} Y_{initial}) \] where \[V_{yfinal} = 0m , Y_{initial} = 0 m/s, V_{yinitial} = (v_0cos(\frac{1}{2} \theta_0))sin\theta_0\] and solve for Yfinal

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much! I can't believe I missed that  especially after writing that equation down! Thank you!

PhysicsGuru
 one year ago
Best ResponseYou've already chosen the best response.1(B) You can find the time it takes to reach max height then multiple by two to get the total time in the air. Once you have the total time you can use the equation for average velocity in the xdirection to find the range: \[V_x = \frac{Range}{Total Time}\]

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much :)

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Wait.. I have to reask this. We could find \(Y_\text{final}\), but not \(\theta_0\)! I have found \(Y_\text{fintal}\).

PhysicsGuru
 one year ago
Best ResponseYou've already chosen the best response.1From the look of the problem setup there are no numbers... it looks like you are finding a formula for \[\theta_0\]

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Right.. 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. I posted a new question, if you want to look there!
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