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Projectile motion

  1. Feb 20, 2006 #1
    A particle is projected on a horizontal ground and moves freely under gravity.
    The horizontal and vertical components of the initial velocity are 2u and u ms^1 respectively.

    The ball lands a distance 80m from the point of projection.

    Show that u=14

    This is what I did:

    Taking motion in the horizontal plane------

    2u = 80/t

    Taking motion in the vertical plane-----

    displacement (s) =0
    acceleration (a)= -9.8
    initial velocity (u) = u
    time of motion (t) = t

    therefore using s=ut +(1/2)at^2

    I get: 4.9t=u

    substituting this into the first equation should give me the correct value of u, but it doesnt. Can someone tell me where i went wrong?
    :confused: :confused:
     
  2. jcsd
  3. Feb 20, 2006 #2
    [tex] 0 = ut - 4.9t^2 [/tex]
    Assuming that t is non-zero, since that is an obvious solution, [tex] t = \frac{2u}{g} [/tex], therefore, since velocity in the horzontial plane is constant, [tex]4u^2 = 80g [/tex], then solve that
     
  4. Feb 20, 2006 #3

    HallsofIvy

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    Why doesn't it? If u= 4.9 t, then t= u/4.9. Putting that into
    2u= 80/t, 2u= (80)(4.9)/u so u2= 40(4.9). What do you think that is?
     
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