A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R. (a) At what angle theta is the rock thrown? (b) In terms of its original range R, what is the range R_max the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.

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A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R. (a) At what angle theta is the rock thrown? (b) In terms of its original range R, what is the range R_max the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.

Physics
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I tried writing equations in such a way that things like v_0 would cancel each other, but I've had no luck. I've been working on it for a while now. Any suggestions?
The maximum height is \[y_{\max} = {v^2 \sin^2 (\theta) \over 2g}\]The range is\[r = {v^2 \over g} \sin(2 \theta)\] We can set these equal together and solve for \(\theta\)
The angle to maximum range can be found to be 45 degrees. This is because 2*45 = 90 which makes sin(2*theta) = 1. This angle is always true regardless of what planet we are on because we are maximizing the sine function and gravity is not part of the sine function.

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What is the derivation for your y_max equation?
I'm just curious as to where it comes from.
I can derive that for you. We know, from kinematics, that the vertical distance can be expressed as\[y = v_y t - {1 \over 2} g t^2\]where \(v_y = v \sin(\theta)\)We need to find the time it takes to reach the maximum height. To do this, we take the derivative with respect to time, set to zero, and solve for t. \[0 = v_y - g t\]\[t = {v_y \over g}\]Substituting this time back into our original equation, we get\[y_\max = v_y \left[ v_y \over g \right ] - {1 \over 2} g \left [ v_y \over g \right]^2\]Simplifying, \[y_{\max} = {v_y^2 \over g} - {1 \over 2}{v_y^2 \over g}\]Combining like terms. \[y_{\max} = {1 \over 2} {v_y^2 \over g}\]Substituting our expression for \(v_y\) in we obtain. \[y_{\max} = {1 \over 2} {v \sin^2(\theta) \over g}\]
Interesting. Thank you! As far as the original problem goes, I'm here, but I'm not sure what to do with the number to solve for theta? \[\sin^2 \Theta =\sin(2 \Theta)\]
Oh, well I'm missing a 2 in there, for one.
Trig identities! I hate 'em but they are great here. \[\sin(2 \theta) = 2\cos(\theta) \sin(\theta)\]\[\sin^2(\theta) = 2 \sin(\theta) \cos(\theta) \rightarrow \sin(\theta) = 2 \cos(\theta) \rightarrow \tan(\theta) = 2\]
I forgot to write a two in the expression above. It should have been... \[\frac{\sin^2 \Theta}{2}=\sin(2 \Theta)\]Doing a similar process to what you did above, I get...\[\sin \Theta = \cos \Theta\]or\[\frac{\sin \Theta}{\cos \Theta}=1\]Is there a good analytical way to solve this? Either by finding at what values sin and cos are equal, or some other method?
Oh. \[\tan \Theta = 1\]
For that, though, I just get 45 degrees, and that's not correct, is it?
That doesn't seem right. Let me solve it by hand.
You are doing your trig identities wrong. \[{\sin^2(\theta) \over 2} = \sin(2 \theta) \rightarrow {\sin^2(\theta) \over 2} = 2 \sin(\theta) \cos(\theta) \rightarrow \tan(\theta) = 4\]
Somewhere around 75 degrees.
Oh! I wasn't doing my trig identities work. I was cancelling out the 2s. Doh. That makes perfect sense. Thank you for your help.
*Wrong.
How did you get the equation for range? I understand taking the derivative of the height function, now I'm confused about the range one...
It is derived in a similar fashion to the maximum height function. Let me walk you through it. We know that there is no acceleration in the horizontal direction, therefore, the range depends on the initial horizontal velocity, \(v_x\) and the time of flight, which comes from the equations of motion in the vertical direction. Let's start with an expression for range. \[r = v \cos(\theta) t \]If we substitute in the time we get for flight time, which is\[t_{flight} = {2 v \sin(\theta) \over g}\]we get\[r = \left ( v \cos(\theta) \right) \left(2 v \sin(\theta) \over g \right)\]\[r = {2v^2 \sin(\theta) \cos(\theta) \over g}\]We can use the following trig identity to further simplify the expression further. \[\sin(2 \theta) = 2\cos(\theta)\sin(\theta)\]Therefore, at last, \[r = {v^2 \sin(2 \theta) \over g}\]
How did you get \[t _{flight}=\left(\begin{matrix}2vsin(\theta) \\ g\end{matrix}\right)\]?
\[t _{flight}\] is twice the time of \[y _{\max}\]

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