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How to make a proof for complementary angles of elevation.

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I have an assignment that is to prove for instance the range of a projectile that is fired from 40 degrees will be the same as one fired from 50 degrees cause they are complementary but I need a general way to prove it for all angles less than 90 degrees.


    2. Relevant equations
    sin θ = cos 90 - θ


    3. The attempt at a solution
    I have searched the Internet and read my physics book and I under stand how to work out if they have the same range or not I'm just having trouble coming up with a general proof for it.
     
  2. jcsd
  3. Aug 1, 2012 #2

    Curious3141

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    The first thing you should start with is to find an expression relating the range R to the angle θ.

    Do you already have an expression, and can you assume it? If not, you have to derive it.
     
  4. Aug 1, 2012 #3
    Ok I'll try.
    No I don't have an expression. I really don't have anything other than what's in my question is all I know really.
     
  5. Aug 1, 2012 #4

    Curious3141

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    Write down the expressions for vertical and horizontal displacement at time t. What should time t be when the particle hits the ground?
     
  6. Aug 1, 2012 #5
    So the vertical = (u sin theta) t
    Horizontal = (u cos theta) t -4.9t^2
    Depending on the angle would depend on what time is when it hits the ground, wouldn't it.?
     
  7. Aug 1, 2012 #6
    Recheck your equations.
    Vertical displacement is zero for the range.
     
  8. Aug 1, 2012 #7
    Ok umm so should
    R = (Vx) (t) With Vx= (Vcos theta) and t= (2*Vsin theta)/g
    So R should = (-V^2 * sin2 theta)/g
    Is that part right?
     
  9. Aug 1, 2012 #8
    Why the minus sign?
     
  10. Aug 1, 2012 #9
    Oops that's not suppose to be there.
    Do you have any suggestions on how I could go about proving that that equation is the same as-
    R= (V^2 * sin 2* 90- theta ) / g
    Also is it suppose to be on gravity or on acceleration?

    Thanks heaps for everyone's help :)
     
  11. Aug 1, 2012 #10
    R= (V^2 * sin 2* 90- theta ) / g is not right and it does not follow from (V^2 * sin2 theta)/g

    Use brackets.
     
  12. Aug 1, 2012 #11
    R= ( V^2 *sin (2* (90- theta)) / g. << is that what you mean by use brackets or is it wrong altogether? :)

    And if it is right if the angles are complementary that R= ( V^2 *sin (2* (90- theta)) / g and R= ( V^2 *sin (2* theta) / g should equal the same range shouldn't they? Or am I completely on the wrong track.? :)
     
  13. Aug 1, 2012 #12
    Recall the unit circle trigonometry and think what sin 2(90 - theta) simplifies to.
     
  14. Aug 1, 2012 #13
    Wrong.
    R=Vxt,
    t=R/Vx ........(1)

    Where you vertical motion equation?
    H=? ..........(2)

    Subtitute (1) in (2) you get the range R.
     
  15. Aug 1, 2012 #14

    Curious3141

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    Why is your gravity term in the horizontal displacement equation? Shouldn't it be in the vertical equation?

    The equations should be:

    [tex]s_y = (u\sin{\theta})t - \frac{1}{2}gt^2[/tex]
    and
    [tex]s_x = (u\cos{\theta})t[/tex]

    where [itex]s_y[/itex] is the vertical displacement and [itex]s_x[/itex] is the horizontal displacement.

    When the particle is at ground level, [itex]s_y = 0[/itex]. Now solve for time t when that happens.
     
  16. Aug 1, 2012 #15

    Curious3141

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    OK, fast forwarding, and assuming you've found R to be given by: [itex]R = \frac{u^2 \sin 2\theta}{g}[/itex], put [itex]\theta = 90^{\circ} - \alpha[/itex]. What is [itex]\sin (180^{\circ} - \beta[/itex]) (where [itex]\beta[/itex] represents any angle)?
     
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