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Help with oblique motion question (initial height)

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    How to calculate initial height (h) in an oblique motion?
    Data: initial velocity (v0) = 10 m/s
    angle (α) = 45º
    g = 10 m/s2
    Variables: x - position in the x axis
    y - position in the y axis
    r - range
    hmax - maximum height
    tf - time of flight
    2. Relevant equations
    vx = vy = 5√2 m/s
    y = h + tan(α) x - (g/v0x2) x2 → trajectory equation
    vy2 = v0y2 - 2gΔy
    3. The attempt at a solution
    1st - 0 = h + v0y tf - ½ g tf2 ⇔ tf = (v0y + √(v0y2 + 2hg))/g
    in the problem tf = (5√2 + √(50+20h))/10
    and using that:
    0 = h + 5√2 × ((5√2 + √(50+20h))/10) - 5×((5√2 + √(50+20h))/10)2 ⇔ 0 = 0 this is redundant becuase I'm using the same formula twice, I understand the result
    2nd - using the trajectory equation:
    0 = h + x - (10/ (2v0x2) x2 ⇔ x = 5 + 5√(1 - (2/5)h)
    in this case x = r so, r = v0x tf ⇒ tf = (1 + √(1 - (2/5)h)/√2
    so I picked the expression from the 1st attempt and matched with this new one, like this:
    (5√2 + √(50+20h))/10 = (1 + √(1 - (2/5)h)/√2 ⇔ h = 0 even though this atempt gave me a "real" result, h ≠ 0 so it can't be
    3rd - in a crazy last atempt I used the expressin for tf from second atempt and substitute it in the equation for y position:
    0 = h + 5√2 × ((1 + √(1 - (2/5)h)/√2 ) - 5((1 + √(1 - (2/5)h)/√2)2 ⇔ h = 0 as expected the result was 0
    4th - I also tried working with torricelli's equation, but that led me no where, I obtained this, which might be helpful: hmax = h + 5/2
     
  2. jcsd
  3. Nov 17, 2014 #2

    Simon Bridge

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    "oblique motion" can refer to a lot of things ... I'm guessing this is a ballistics problem?
    i.e. a projectile is fired at speed 10m/s at an angle 45deg to the horizontal - neglect air resistance.
    Is that correct?

    If so then
    ... this is incorrect.
     
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