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Am I analizing this correctly?

  1. Dec 9, 2006 #1
    Hi!
    I was given this physics problem and I'm trying to solve it, but I'm not sure if I'm doing it correctly...
    I have to write a formula to calculate the _minimal_ speed (I missed this part at first) at which a projectile must be shot to pass through point (x, y).

    My first step was to write a formula to calculate the exact speed at which a projectile must be shot to pass through point (x, y) when shot at an angle 'a'.
    This is what I get:

    s = (sqrt(g) * x) / (sqrt(2 * (x * tan(a) - y)) * cos(a))
    (g >= 0, a != 90, a > arctan(y / x))

    I'm quite sure that it is correct.
    Now I need to figure out how to find the formula for the _minimal_ speed.
    I'm quite sure that it depends on the angle.

    sqrt(2 * (x * tan(a) - y)) * cos(a))
    The bigger this part gets, the smaller the speed gets.
    '2', 'x' and 'y' are constant, so it must be 'a', right?
    So I need to find the angle at which both 'tan(a)' and 'cos(a)' give the biggest values.
    How do I do that?
    Am I thinking correctly?
    Does any of this actually make any sense?
     
  2. jcsd
  3. Dec 9, 2006 #2

    arildno

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    I'll look a bit closer later on, but your reasoning seems sound; now consider:

    Let us assume you have correctly found the speed needed as a function of "a" (I haven't checked that yet).

    How can you find where a function has its minimum value?
     
  4. Dec 9, 2006 #3
    How should I know?
    I'm a 9th grade student... (Who really sucks at math!)

    [EDIT]

    The only way I can think of is using a graph, but that doesn't help me to write a formula...
     
    Last edited: Dec 9, 2006
  5. Dec 9, 2006 #4
    Does this help?:
    For MINIMAL speed, you will probably want the top of the projectiles arc to go through your (x,y) point.
     
  6. Dec 9, 2006 #5

    arildno

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    Have you learnt to differentiate functions?
     
  7. Dec 9, 2006 #6
    Nope.
    Anyway, according to multiple graphs, the optimal angle is somewhere near 0.8 radians, that's somewhere near 45 degrees...
    The graphs are, of course, inaccurate, but I bet it is exactly 45 degrees.
    I thought so from the very beginning, but it seems kind of odd...
    Some points can't be reached at all from this angle...
    Could this actually be correct?

    [EDIT]

    The 'x' doesn't seem to have any effect but a big change of 'y' changes the angle...
    Ok, now I officially give up...
     
    Last edited: Dec 9, 2006
  8. Dec 9, 2006 #7

    arildno

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    So are you SURE you don't know what a derivative is?
     
  9. Dec 9, 2006 #8
    Nope, but let's assume that I do.
    How does that help me?

    [EDIT]

    Isn't there a simple way to solve this problem??

    [EDIT 2]

    This is actually a parabola...
    There must be a formula to find the minimum of a parabola, right...?
     
    Last edited: Dec 9, 2006
  10. Dec 9, 2006 #9

    vanesch

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    In 9th grade, that's not so surprising, no ?
     
  11. Dec 9, 2006 #10

    arildno

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    I don't know what 9'th grade compares to in Norway.
     
  12. Dec 9, 2006 #11

    arildno

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    Well, from my calculations, I get that we should have:
    [tex]\tan(a)=\frac{y+\sqrt{x^{2}+y^{2}}}{x}[/tex], where I have assumed that both x and y is greater than zero, and the projectile was shot up from the origin.
    I'll see if I can make a neat geometrical justification of this for you.
     
    Last edited: Dec 9, 2006
  13. Dec 9, 2006 #12
    Is 'a' the angle of minimal speed?
     
  14. Dec 9, 2006 #13

    arildno

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    Yes, "a" would be the angle for minimal speed.
     
  15. Dec 9, 2006 #14
    I'll check if it works.
    Thanks. :)
     
  16. Dec 9, 2006 #15

    arildno

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    Please observe the last-second edit I made to the formula!
     
  17. Dec 9, 2006 #16
    Yep. It corresponds to the results from the graphs.
    Thanks!
    But how did you come up with it?
     
  18. Dec 9, 2006 #17

    arildno

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    With differentiation techniques..
     
  19. Dec 9, 2006 #18
    Could you please post your calculations?
    I'll try to figure them out with this calculus book I found on my dad's bookshelf...
    (Yes, I WILL figure it out... After all, they don't teach us trigonometry in 9th grade either and I did figure that out...)
     
  20. Dec 9, 2006 #19

    arildno

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    1. First of all, I assume you will agree that if we find the angle value yielding the minimum of the squared speed, then we have found the angle value that minimizes the speed itself.

    2. Secondly, we have the identity, for any angle a:
    [tex]\tan^{2}(a)+1=\frac{\sin^{2}(a)}{\cos^{2}(a)}+\frac{\cos^{2}(a)}{\cos^{2}(a)}=\frac{1}{\cos^{2}(a)}[/tex]

    3. Thus, using "s" for speed, we have the formula:
    [tex]s^{2}=\frac{gx}{2}\frac{\tan^{2}(a)+1}{\tan(a)-\tan\beta}, \tan\beta=\frac{y}{x}[/tex]

    Agreed?
    (It is just a rewriting of your own formula)

    I'll proceed later on..
     
  21. Dec 9, 2006 #20

    arildno

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    Note therefore that we only need to find the value of tan(a) that minimizes the partial expression [tex]\frac{\tan^{2}(a)+1}{\tan(a)-\tan\beta}[/tex]
    Agreed?
     
    Last edited: Dec 9, 2006
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