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Projectile Motion- What angle does range equal max height?

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data

    At what projection angle will the range of a projectile equal its maximum height?

    2. Relevant equations

    Big 5

    vf2= vi2 + 2ad

    3. The attempt at a solution

    So I attempted to solve this question by randomly subbing in equations into each other.

    If i draw out a diagram to solve for the angle, I would need the x and y components.

    Let V equal the total velocity.

    viy= Vsinθ
    Plugging into vfy2= viy2 + 2aydy , I get

    vf2= V2sin2θ + 2aydy

    Now I isolate for d, and end up with dy= (V2sin2θ)/2ay

    Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

    Now using the range formula, I can determine the horizontal displacement.

    dx= (V2sin2θ)/g

    (g=ay)
    With these two formulas, I substitute and...

    (V2sin2θ)/2ay = (V2sin2θ)/ay

    sin2θ = aysin2θ

    sin2θ = ay(2sinθcosθ)

    sinθ = 2aycosθ

    sinθ/cosθ = 2ay

    tanθ = 2(-9.8)

    θ = tan-1(-19.6)

    θ = 87°

    However... this is not the answer that the answer key states. The answer shown is 76°, and I do not know where I have gone wrong, or if I am doing this right.

    Sorry if it was a mess and confusing, thanks for the help!
     
  2. jcsd
  3. Feb 26, 2014 #2
    No. That should be $$ d_y = - \frac {v^2 \sin^2 \theta} {2a_y} $$ Note the minus sign.

    Assuming the above is correct (which it is, because you made another sign error there!), how do you get this:

    ?
     
  4. Feb 26, 2014 #3
    I got sin2θ = aysin2θ

    from

    (V2sin2θ)/2ay = (V2sin2θ)/ay

    Cross multiply and cancel out one of the ay

    ay(V2sin2θ) = 2ay(V2sin2θ)

    [STRIKE]ay[/STRIKE] (V2sin2θ) = [STRIKE]2 [/STRIKE]ay(V2sin2θ)

    Divide both sides by V2

    [STRIKE]V2[/STRIKE]sin2θ = ay([STRIKE]V2[/STRIKE]sin2θ)

    sin2θ = aysin2θ
     
  5. Feb 26, 2014 #4
    You need to brush up on your algebraic manipulation skill. You do not cancel "one of the a" like that.
     
  6. Feb 26, 2014 #5
    If one side has "2a" and the other has "a", can't you divide out one of the "a" by dividing both sides by it?
     
  7. Feb 26, 2014 #6
  8. Feb 26, 2014 #7
    If you divide both sides by "a", each side gets divided by "a". So if one side has "2a", and the has "a", how can the result of that be "a" and "1"? Do you really think that the inverse operation, "a" times "a", is "2a"?
     
  9. Feb 26, 2014 #8
    Ok I see what you mean. I'm left with 2 instead of 'a'. Mistakened as a square
     
  10. Feb 26, 2014 #9
    So what do you get now?
     
  11. Feb 26, 2014 #10
    Thanks for the help :). After correcting that I end up with 76° :D
     
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