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Ratio of flight times at angles 45degree +/- alpha

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider two projectiles launched on level ground with the same speed,
    at [itex]angles 45°\pm \alpha[/itex].
    Show that the ratio of their flight times is [itex]tan 45°\pm \alpha[/itex].

    3. The attempt at a solution

    projectile 1 launched at [itex] Θ = 45° + \alpha[/itex]:

    vyf1 = vyi1 + gt
    tapex1 = vyi/g
    ttotal flight time1 = 2vyi1/g
    = 2 vi sin (45 ° + [itex]\alpha[/itex])


    projectile 2 launched at [itex] Θ = 45° - \alpha[/itex]:

    vyf2 = vyi2 + gt
    tapex2 = vyi2/g
    ttotal flight time2 = 2vyi2/g
    = = 2 vi sin (45 ° - [itex]\alpha[/itex])
     
  2. jcsd
  3. Jan 25, 2014 #2

    BvU

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    Correct. Now write out the trigonometry!
     
  4. Jan 25, 2014 #3
    I'm not sure it's correct. I mean, intuitively, tan = sin / cos.

    Both of my equation for the flight time are in sin.
     
  5. Jan 25, 2014 #4
    What do you mean by writing out the trigonometry?
     
  6. Jan 25, 2014 #5
    Incidentally, the given answer is incorrect. It should be [itex](tan(45) \pm2α)=1\pm2α[/itex], and this only applies in the limit of small values of α.

    You first take the ratio of the two flight times, and then apply the trig formulas for the sum and difference of two angles. Then you linearize with respect to α.
     
  7. Jan 25, 2014 #6

    gneill

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    I don't think that the given answer is incorrect. ##tan(\frac{\pi}{4} + \alpha)## looks okay to me...
     
  8. Jan 25, 2014 #7
    Wow. Very cute. I must admit, I didn't think of putting the cosine of the complement in the denominator. Well, at least I got the right answer in the limit of small α. Not very satisfying for me, however.

    Chet
     
  9. Jan 25, 2014 #8

    BvU

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    Gentlemen, gentlemen: tan45°±α means tan (45°±α). Of course (right, negation?). And it is not a linearization, approximation or anything. It has to do with the funny trigonometric properties of [itex]\frac{\pi}{4}[/itex]. Write it out !

    Your are correct, but: intuition? If you don't believe it, look it up (or make a drawing ;-) )

    Your equations are in sin, yes, they are. The ratio of the flight times is sin (45 ° + α) / sin (45 ° - α). That are sines of sum and difference, which can be written out (as the hint suggested). Something physicists and Excel users look down on, but it can be done.

    Here we go: using

    [itex]\sin(\frac{\pi}{4}+\alpha) = \sin(\frac{\pi}{4}) \cos\alpha + \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha + \sin\alpha \right )
    [/itex] and

    [itex]\sin(\frac{\pi}{4}-\alpha) = \sin(\frac{\pi}{4}) \cos\alpha - \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha - \sin\alpha \right )
    [/itex] to get a ratio

    [itex]\frac{ \cos\alpha + \sin\alpha}{ \cos\alpha - \sin\alpha}[/itex]

    (sorry, it's been over 25 years since I last used TeX and it's a little rusty now, don't know how to get it in comparable font sizes any more)

    Anyway, now divide by [itex]\cos\alpha[/itex] above and below to get a ratio

    [itex]\frac{ 1+ \tan\alpha}{ 1- \tan\alpha} =\frac{ \tan\frac{\pi}{4} + \tan\alpha}{ 1-\tan\frac{\pi}{4} \tan\alpha} [/itex]

    If we remember (or look up, as I had to do)

    [itex]\tan(\gamma+\alpha) = \frac{\tan\gamma + \tan\alpha} {1- \tan\gamma \tan\alpha}[/itex] we see that they are the same !
     
  10. Jan 25, 2014 #9
    Gneill implied something much more elegant:

    [tex]\frac{\sin(\frac{π}{4}+α)}{\sin(\frac{π}{4}-α)}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{2}-(\frac{π}{4}-α))}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{4}+α)}=\tan(\frac{π}{4}+α)[/tex]
     
  11. Jan 25, 2014 #10

    BvU

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    I love it! Missed the implication because I had it more or less worked out last night and was too sleepy to catch that. Anyway, dusting off old sum rules was a good thing.
    Working back from the answer would have increased the chance of finding this very short proof...
    And doodling with a plot of sin(x) and cos(x) probably also would have helped....
     
  12. Jan 25, 2014 #11

    gneill

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    :smile: Good advice! Problem solving epiphanies often spring from dredging up the basic trig function relationships and identities. All those courses one thought done and dusted in the prerequisites course list regularly turn out to be a goldmine for these "aha!" moments.
     
  13. Jan 26, 2014 #12

    BvU

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    I'm not sure negation has a satisfied feeling about all this help going over his head. How is it ?
     
  14. Jan 26, 2014 #13
    I must admit to the elegance and simplicity of the above equation. It's a good refresher. I had the time of flight in symbolic form for both particles fired at an arbitrary deviation of +/- alpha but couldn't reduce the ratio to the intended identity.
     
  15. Jan 26, 2014 #14
    I just got home from work. I'll be looking it through after taking a shower.
     
  16. Jan 26, 2014 #15

    Can I not multiply by cos(a) + sin(a)?
     
  17. Jan 26, 2014 #16

    BvU

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    Would it help ?
     
  18. Jan 26, 2014 #17
    You're right. It's not an appropriate step to arriving at the intended identity.
     
  19. Jan 26, 2014 #18
    I understand how 1+ tan(a) is tautological to tan(pi/4) + tan(a)
    but how does 1 - tan(a) = 1 - tan(pi/4) tan (a)?
     
  20. Jan 26, 2014 #19

    BvU

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    tan(pi/4) = 1, so you can replace a 1 with this guy anywhere you want...
     
  21. Jan 27, 2014 #20
    I know and which I did.

    However, 1 + tan (a) : 1 - tan(a) = tan (pi/4 + a) : tan (pi/4 - a)

    gives me [tan(pi/4) + tan(a)]/1 - tan(pi/4)tan(a) : [tan(pi/4) - tan(a)]/1 + tan(pi/4)tan(a)

    I'm unable to reduce it further.
     
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