What is the optimal angle for maximum distance in a shot put throw?

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Homework Help Overview

The discussion revolves around determining the optimal angle for maximum distance in a shot put throw, given an initial speed and release height. The subject area includes projectile motion and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for finding the optimal angle, including differentiation of the range formula and intuitive reasoning. Questions about the appropriateness of deriving the range formula in a high school context are also raised.

Discussion Status

The discussion is active, with various approaches being suggested. Some participants offer mathematical insights while others express caution about the complexity of the derivation. There is no explicit consensus on the best method to find the optimal angle.

Contextual Notes

Participants note the original poster's preference for guidance without direct solutions, indicating a focus on understanding rather than simply obtaining answers. There is uncertainty regarding the level of mathematical derivation expected in the context of high school physics.

Asteldoth
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Bill threw a shot put with an initial speed of 12m/s at an angle of 35 degress to the horizontal. He released the shot put at a height of 1.5 meters. How far did he throw the shot put? If he could have changed the angle of his throw, at what angle would he had attained the maximum distance, and would would that distance have been?


I got the how far question, but I'm confused on how to find the maxium distance and angle. Anyone have any tips? Please don't solve it for me (I don't know if you guys do that here, but I'd like to figure that out on my own), I just need to know how to find it.
 
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Solve it with some variable angle theta and differentiate the function to find max and mins. Alternatively, as that might get a bit messy with inverse trig functions everywhere, think about it intuitively, as this answer is one you can get just by applying logic, which is handy :)
 
There is a vector approach that I think makes the optimal angle pretty obvious. Do you know the range formula? It can be derived like so:
[tex]y=(v_0 sin \alpha)t -\frac{1}{2} g t^2[/tex]
[tex]\frac{dy}{dt} = v_0 sin \alpha - gt[/tex]
set the derivative to zero to find the maximum height, and then solve for time
[tex]t = \frac{v_0 sin(\alpha)}{g}[/tex]
[tex]x=v_0 (cos (\alpha))t[/tex]
Plug in t to the equation.
[tex]R = v_0 (cos (\alpha)) \frac{v_0 sin(\alpha)}{g}[/tex]
Use a double angle identity and you finally get:
[tex]R = \frac{v_{0}^{2}}{g} sin2 \alpha[/tex]

You probably have been given this formula in some form or another. So just stare at for a little and find out which angle would make that equation the biggest.
 
well he said he didnt want the answer so i tried to be as elusive as possible, but anyway, good answer
 
I didn't know if deriving the range formula was in the scope of a high school physics class. It could be done, but wouldn't be completely obvious. Asteldoth, if you didn't want that much help then sorry, but looking back I actually made a small error that you can see if you can catch (follow the steps). Also, you still have to see what angle is optimal.
 

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