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Projectile motion  max range 
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#1
Sep1011, 11:41 AM

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1. The problem statement, all variables and given/known data
After i get the formula x = Vo^{2}sin(2θ)/g, I was told that I can take the derivative of x and let that equal to zero to get the max range of the projectile. Why? What does taking the derivative do in order to help us find the max angle? I know that the value of the derivative at the maximum height of the traj. would be 0, but why is that significant? 2. Relevant equations x = Vo^{2}sin(2θ)/g => dx/dθ = Vo^{2}/g * 2cos2θ 


#2
Sep1011, 12:02 PM

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A function goes through an extremum (maximum or minimum) when its derivative is zero. Positive derivative means that the function is increasing, negative derivative means that the function is decreasing. Therefore, zero derivative means that it is neither increasing nor decreasing, i.e. it is at an extremum.



#3
Sep1011, 12:11 PM

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I already knew that. My question is, how is that relevant to how we get the optimal angle for max range?



#4
Sep1011, 12:14 PM

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Projectile motion  max range
You could just look at the graph of x = Vo^{2}sin(2θ)/g, and visually determine what angle causes an x maximum, but this is technically an imprecise method (if this function was not so darned symetrical, you could see where eyeball error might come into play). But if you set the derivative to 0 and solve for θ, you can mathematically determine exactly what angle causes the x maximum.



#5
Sep1011, 12:18 PM

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#6
Sep1011, 12:39 PM

P: 8

Okay, I know what confuses me now. Before, I thought the path of the projectile is what I'm supposed to be looking at. Instead, I should be looking at the graph of the relationship between θ and x. At θ=pi/4, the maximum value of x is achieved. If you only look at the path of the projectile, that would be rather obscure. And by taking the derivative of x, you get a line with a negative slope, where the zero is the value at which x is at max (the extremum of the x function)... Is that correct?



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