1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile: launching and landing from the same height help me ?

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    If an athlete jumped 2 feet high and left the ground at an angle of 20 degrees with respect to the horizontal, how fast was the athlete going in the forward (positive horizontal) and upward (positive vertical) directions immediately after takeoff? If the height of takeoff was the same as the height of landing, how fast was the athlete going in the horizontal and vertical directions right before landing?

    2. Relevant equations

    vx = vcos(thata)
    vy = vsin(theta)

    Immediately after take off :
    vx = 2cos25 = 1.8126
    vy = 2sin25 = 0.845

    right before landing:

    vx = 2cos25 = 1.8126
    vy = 2sin25 = -0.845


    3. The attempt at a solution


    Im not sure if my solution is correct can anyone confirm please ?
     
  2. jcsd
  3. Nov 22, 2007 #2

    rl.bhat

    User Avatar
    Homework Helper

    Angle is 20 not 25. What is the equation for h?
     
  4. Nov 22, 2007 #3
    Sorry i put the wrong theta i took the theta for another problem.

    The equation for H is as follows:

    [​IMG]

    So basically what i think its going to be the 2nd one from the Y- Direction Equations?
     
  5. Nov 22, 2007 #4

    rl.bhat

    User Avatar
    Homework Helper

    You can use third one. In that Vfy = 0.
     
  6. Nov 22, 2007 #5
    If i use the third one i will have one unknown which is d

    0 = viy^2 + 2ady

    dy = -viy^2/2a


    dy = -(2sin20)^2/2(32)

    dy = 7.3 * 10 ^ 03 ??
     
  7. Nov 22, 2007 #6

    rl.bhat

    User Avatar
    Homework Helper

    Here d=2 feet not vi
     
  8. Nov 22, 2007 #7
    THANKSSS

    0 = viy^2 + 2ady

    Rearranging :
    Viy^2 = -2ady


    viy^2 = -2(-32)(2)
    viy = 11.3 m/s

    At take off:
    Vv= 11.3 m/s

    At landing:
    Vv = -11.3 m/s


    Heres the solution after writing it down , all coments are welcomed:

    [​IMG]
     
    Last edited: Nov 22, 2007
  9. Nov 22, 2007 #8

    rl.bhat

    User Avatar
    Homework Helper

    Good!!
     
  10. Nov 22, 2007 #9
    All Credits go to you thanx a lot
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?