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How to find initial velocity without being given time

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    a projectile is launched at an angle of 26.5 above the horizontal at a height of 1.11m. the range is 2.331m. find the initial velocity and time given that acceleration due to gravity is 9.8m/s^2.

    3. The attempt at a solution
    first, I split the velocity into its components:

    v1x = v1cos26.5
    v1y = v1cos26.5

    then,
    v1x = dx/t
    v1cos26.5 = dx/t
    rearranged to isolate for time since it is not given:
    t = dx/ v1cos26.5

    also, using the kinematic equation: v2y^2 = v1y^2 + 2at
    and since at the projectile's max, v2y = 0m/s and since the max occurs at t*(0.5) ***(but I think this is inaccurate because the projectile does not land at the same height in which it was released)

    so,
    v2y^2 = v1y^2 + 2at
    0 = v1y^2 + 2(9.8)(t/2)

    then isolate for t again,

    t = 2(v1y)^2 / 4.9
    or
    t = 2(v1sin26.5)^2 / 4.9

    then make each isolated time equation equal one another:

    dx/v1cos26.5 = 2(v1sin26.5)^2 / 4.9

    then isolate for v1:
    v1 = sqrt( 4.9*dx / (sin26.5cos26.5) )
    since dx = 2.331,
    v1 = sqrt( 4.9*2.331/ (sin26.5cos26.5) )
    v1= 5.35m/s

    Therefore, v1 = 5.35m/s

    then for time,
    v1x = dx/t
    t = dx/v1cos26.5
    = 2.331 / 5.35cos(26.5)
    = 0.487s

    Therefore, t = 0.49s
     
  2. jcsd
  3. Feb 10, 2015 #2

    NascentOxygen

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    Staff: Mentor

    Hi Qwerty123!. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    The correct equation is v2 = u2 + 2as :(

    See how you go now.
     
    Last edited by a moderator: May 7, 2017
  4. Feb 11, 2015 #3
    Thanks for replying. :). I'm not familiar with the variables u and s?
     
  5. Feb 11, 2015 #4

    gneill

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    Staff: Mentor

    It's a common convention to let u be the initial velocity and s represent distance ("s"pace I suppose).
     
  6. Feb 11, 2015 #5

    NascentOxygen

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    Staff: Mentor

    Maybe you'd like it as: v2y^2 = v1y^2 + 2.a.d
    where that last term involves the y component of acceleration times distance in y direction.

    Check your course notes.
     
  7. Feb 12, 2015 #6
    This actually varies between countries and even states. For example in Victoria, Australia, x is used for distances. However Queensland uses s.

    I can't remember where I saw it but there's a webpage out there that lists the most common ones.
     
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