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Projectile Jumper Problem

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A long jumper leaves the ground at 45 degrees above the horizontal and lands 8.2 meters away. What is her "takeoff" speed v0?

    2. Relevant equations

    x = x0 + v0t + 1/2at^2
    v^2 = v0^2 + 2ad
    v = v0 + at

    3. The attempt at a solution

    I can't seem to figure this problem out. Here's the information I got from the problem so far:

    x component:

    x0 = 0
    x = 8.2 m
    a = 0

    y component:

    y0 = 0
    a = -9.8 m/s^2

    There seems to be too many missing variables to solve the equation. I even tried breaking the projectile in half, which yielded me this:

    x component:

    x0 = 0
    x = 4.1 m
    a = 0
    v = v0

    y0 = 0
    a = -9.8 m/s^2
    v = 0

    EDIT: I forgot to mention I also realize that v0x and v0y components are equal, since the angle is 45 degrees. I know this is important, but I can't connect it to anything.

    I still can't seem to figure it out, since I need y, or how high the object goes at its highest point. I also can't figure out the time it takes... If I knew either of those, I could figure it out, but I can't seem to get anything from this.

    Thanks for your help.
    Last edited: Sep 13, 2008
  2. jcsd
  3. Sep 13, 2008 #2
    A few points you may find useful, for the full length problem

    1) the jumper 'lands' - this means that at the end of the jump the y displacement is zero as she returns to the ground (y = 0)

    2) the point above means that the final velocity in the y-direction must be equal in magnitude to the initial velocity but in the opposite direction (vy = -v0y)

    3) you can solve for the time in the y-direction using v = v0 + at

    I think you should be able to manage the rest. The trick is just to realize what's happened in the y-direction at the end of the jump.
  4. Sep 13, 2008 #3
    i understand that y=0 and vy=-v0y...

    what i dont understand is how you can use v = v0 + at to find the time, since you still don't know what v0 (or v) is.
  5. Sep 13, 2008 #4
    I should just mention that time will be a function of the v0y. When you use this time in the equation for the x-direction velocity you'll end up with something like v0x * v0y = number. Then just use what you know about the angles to get the answer.
  6. Sep 13, 2008 #5
    I solved it!

    It wound up being 9.0 m/s, after I found the v0x=v0y=6.338769597

    Thanks so much for your help!
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