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Projectile Motion High Jumper

  1. Apr 8, 2009 #1
    Physics Kinematics Unit: Question concering Projectile Motion

    1. The problem statement, all variables and given/known data
    If a high jumper reaches her maximum height as she travels across the bar, determine the initial velocity she must have to clear a bar set at 2.0 m if her range during the jump is 2.0 m. What assumptions did you make to complete the calculations?

    Range(d) = 2.0 m
    Acceleration due to gravity (a) = 9.81 m/s^2 down
    The height of the bar = 2.0 m

    2. Relevant equations

    [tex] a= \frac{Vf - Vi}{t} [/tex]

    [tex] d=(Vi)(t) + (1/2)(a)(t^2) [/tex]

    [tex] d=Vt [/tex]

    3. The attempt at a solution
    This is my attempt. Since Vix is O, we can determine time using this equation:
    [tex] d=(Vi)(t) + (1/2)(a)(t^2) [/tex]

    [tex] 2.0 m = (0)(t) + (1/2) (9.81 m/s^2) (t^2)[/tex]

    [tex] t = 0.64 s[/tex]

    Now trying to find Vix, use this equation:

    [tex] d = vt[/tex]

    [tex] 2.0 m = V(0.64) [/tex]

    [tex] v = 3.13 \frac {m}{s} [/tex]

    Now You have to divide the time by 2, because that's half way through the jump, her maximum height would be half the time.

    [tex] t = \frac {0.64}{2} [/tex]

    [tex] t = 0.32 s [/tex]

    Now find Viy, using this equation:

    [tex] d = Vt[/tex]

    [tex] 2.0 = V(0.32)[/tex]

    [tex] v = 6.2 \frac {m}{s}[/tex]

    Now find Vi using this this equation:

    [tex] a^2 + b^2 = c^2 [/tex]
    [tex] 6.25^2 + 3.13 ^2 = c^2[/tex]
    [tex]c= 6.99 m/s (63 degrees)[/tex]

    I got this wrong, but I'm just showing you my attempt to this question
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2


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    Homework Helper

    Motion of the high jumper is the projectile motion.
    So vyi = vsin(theta) and vxi = vcos(theta)
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