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Projectile Motion diver leap

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A diver leaps from the top of a tower with an initial horizontal velocity of 5.0 m/s and an upward velocity of 2.0 m/s. Use a vector diagram to indicate the diver's horizontal and vertical velocity components 1.0s after he leaves the tower.

    2. Relevant equations
    y= vyt + 1/2gt2
    maybe v=d/t

    3. The attempt at a solution

    I'm not exactly sure how to go about this problem. Do I just use the x and y formula, and that is all?
    Vx= x/t

    Vy= (y-1/2gt2)/t

    Would those be right?
  2. jcsd
  3. Nov 26, 2008 #2
    Not quite right. The best equation to use here is v2 - v1 = a(t2 - t1). So for vx, a =0 and v2 = v1 = 5. (as you said). For vy, a = -9.8 and v2 = 2 - 9.8 = -7.8.
  4. Nov 26, 2008 #3
    Really? They haven't mentioned that formula in what I've done so far...
  5. Nov 27, 2008 #4


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    Homework Helper

    I had never witnessed such a formula either till it clicked - it's just a slight modification to the [tex]v=u+at[/tex] formula.
    v = final velocity
    u = initial velocity
    a = acceleration due to gravity
    t = time

    And yes, this is the best formula to use in this case. Horizontal velocity will be kept constant at 5ms-1 since we are assuming no resistances, so you already know what that vector is going to be after 1 second. Now you just need to calculate the vertical vector with this equation, which ak1948 already gave as -7.8ms-1.
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