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Projectile Motion Jump Question

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A person jumps 8.9m. The person's takeoff speed was 15m/s. What is the angle relative to the ground at takeoff?


    2. Relevant equations



    3. The attempt at a solution

    I found t using the kinematic formula to be 0.5s using the formula. d=vit+1/2at^2. Can someone tell explain to me what step to do next. Thanks
     
  2. jcsd
  3. Mar 25, 2014 #2
    I assume the jump distance stated is horizontal only.

    You can't use that kinematic formula straight out the way you did. You have to break it into two different directions. So you will have two formulas but three unknowns: t, dy, and θ. You get the 3rd equation using a trigonometric or geometric equation.
     
  4. Mar 25, 2014 #3
    ummm, i attempted it so basically i used y=1/2at^2. so imagine you're at the peak of 8.9 meters, use that equation to calculate the amount of air time before reaching the air. after that use that t to plug in the vertical kinematic equation:
    y= y(knot) + v(knot)t - 1/2at^2. plug 8.9 in. For v(knot) set it equals to v(knot)sin(theta). Then just find theta.
    it's gonna be parabolic path, therefore the time it takes to get in to the air is the same as the time it takes to fall.

    I'm assuming that he's jumping 8.9 vertically?so the math I just articulated is for vertical distance
     
    Last edited: Mar 25, 2014
  5. Mar 25, 2014 #4
    Ummm...well you have to clarify is the person jumping 8.9m high (y direction) or 8.9m horizontally (x direction)?
     
  6. Mar 25, 2014 #5
    I'm sorry about that. 8.9m was for the horizontal distance.
     
  7. Mar 25, 2014 #6

    SteamKing

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    If someone could jump 8.9 m vertically, he would be the president for life of Phi Slamma Jamma.
     
  8. Mar 25, 2014 #7
    Right, so then you can't use the equation the way you did.

    Can you set up the 3 different equations?
     
  9. Mar 26, 2014 #8
    ok it's pretty simple, you set up 2 equations of T(air time)
    so the first one is t=(vsin(theta))/g and they other one is t=x/(vcos(theta))
    set them equal to each other, plug v and x in and find theta, that's it

    Edit: sry, for the first one just times it by 2 so t=2(whatever is there)
     
    Last edited: Mar 26, 2014
  10. Mar 26, 2014 #9
    You'll need one more trigonometric identity equation to find theta.
     
  11. Mar 26, 2014 #10
    Equation for trajectory says $$x = \frac {{v_{0}}^2\sin 2\theta}{g}$$
    Plug in your unknowns from there.
     
  12. Mar 26, 2014 #11
    Thanks guys for the help!
     
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