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

AI Thread Summary
The discussion centers on calculating the maximum distance of a shot put throw given an initial speed of 12 m/s at a 35-degree angle and a release height of 1.5 meters. Participants suggest using the range formula, R = (v_0^2/g) * sin(2α), to determine the optimal angle for maximum distance. They emphasize the importance of differentiating the function to find maximum values and suggest an intuitive approach to understanding the problem. The conversation also touches on the complexity of deriving the range formula and its applicability in high school physics. Ultimately, the optimal angle for maximum distance in projectile motion is typically 45 degrees, but the discussion encourages independent problem-solving.
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:
y=(v_0 sin \alpha)t -\frac{1}{2} g t^2
\frac{dy}{dt} = v_0 sin \alpha - gt
set the derivative to zero to find the maximum height, and then solve for time
t = \frac{v_0 sin(\alpha)}{g}
x=v_0 (cos (\alpha))t
Plug in t to the equation.
R = v_0 (cos (\alpha)) \frac{v_0 sin(\alpha)}{g}
Use a double angle identity and you finally get:
R = \frac{v_{0}^{2}}{g} sin2 \alpha

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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