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Answer check-derive a general formula for horizontal displacement

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

    Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed vi from height, h.


    3. The attempt at a solution

    vyf = vyi -gt
    t = vyf/g
    x = vxi .t = vi cosΘ (t = vyf/g) = [vi^2 sinΘcosΘ]/g
     
  2. jcsd
  3. Jan 25, 2014 #2

    BvU

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    The launch is horizontal, so your theta is known. You run into trouble with sin theta, which is 0.

    Also, I don't find any dependence on h. Don't you think that if you launch from a high fortress, the projectile covers a bigger distance than if you launch from a small hill ?

    Tip: make a drawing!
     
  4. Jan 25, 2014 #3
    My solution was to find the time taken for the projectile to reach the ground. Then sub that time, t, into the horizontal displacement formula vi cos Θ .t. Is there a reason why this is not viable?
     
  5. Jan 25, 2014 #4

    BvU

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    The idea is excellent. But don't you think that h should appear somewhere in that time ?
    You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
    because vyf = gt.

    Made a drawing yet ?
     
  6. Jan 25, 2014 #5
    I have the drawing already. It's really just the second half of the parabola.
     
  7. Jan 25, 2014 #6
    It's not an issue, is it? Since Θ = 0°, then, vi sin(0°) = 0.
     
  8. Jan 25, 2014 #7

    BvU

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    Two approaches possible, in fact they are the same.
    You can get vyf from an energy balance mgh = 1/2 m vyf^2
    or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.
     
  9. Jan 25, 2014 #8
    It struck me that the equation y(t)= y(0) + vi.t -0.5gt^2 can be reduced to y(t) - y(0) = h = vi.t - 0.5gt^2. From there I can solve for time. I think this approach is a parallel reasoning to the one in my attempted solution and hence I do not think my answer in the OP is wrong. The only difference would be that using h allow me to express the solution in the intended manner the question demands.
     
  10. Jan 25, 2014 #9

    h = vyi.t - 0.5gt^2
    but since yf < yi , then, yf - yi = -h, and vyi = 0 since vi sin 0° = 0
    hence, - h = -0.5gt^2
    t = SQRT[2h/g] (time taken for projectile to reach the ground)

    given time, t = SQRT [ 2h/g], the horizontal displace in this period of time
    = vxi.t = vi cos(0°) [SQRT[2h/g]] = vi[SQRT[2h/g]]
     
  11. Jan 25, 2014 #10

    BvU

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    Excellent.
     
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