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Avg speed

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A soccer ball is kicked from the ground with an initial speed of 19.8 m/s at an upward angle of 45°. A player 55 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to catch the ball just before it hits the ground?

    2. Relevant equations

    x(t)= initial position + final velocity * time

    v(t)= (acceleration * time) + initial velocity

    x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

    x= initial position * (average velocity * time)

    average velocity= (final velocity - initial velocity) / (2)

    (final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position

    3. The attempt at a solution

    I came out with 11.19 but it says its wrong??
  2. jcsd
  3. Feb 13, 2009 #2
    try 24.73 m/s if its right ill explain
  4. Feb 13, 2009 #3


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    First of all how far away does the ball land? That determines how far Cristiano Ronaldo has to run.

    Then you need to determine how long the ball is in the air.

    So what did you get for those?
  5. Feb 13, 2009 #4
    heres my explanation anyways, it says its at 19.8 m/s at that angle, therefore the x and y components of velocity respectively will be 19.8cos45 and 19.8sin45. for the y compenent of this solve for the time the ball is in the air. that is use vf = vi +aT. therefore the time is the amount of time it is in the air, using this time and the x component of the velocity, solve for how far it goes.
    now u have the x distance the ball travels. with both that distance and how far away the person is, u can solve the velocity at which the person must run to meet with the ball to catch it.
  6. Feb 13, 2009 #5


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    Not even Theo Wolcott runs that fast.
  7. Feb 13, 2009 #6
    I did it once more and came out with 6.73. Still wrong
  8. Feb 13, 2009 #7
    prove it
  9. Feb 13, 2009 #8
    I solved for the y component and that came out to 14 m/s. I don't understand what to do with the information though. How do I solve for time using just the y component?
  10. Feb 13, 2009 #9
    k my bad the answer is 5.36 ur welcome
  11. Feb 13, 2009 #10
    okay, you have 14 m/s for the y component, just imagine shooting a ball straight into the air, perpendicular with the ground, wtih 14 m/s. now, u have 14 as Vi (velocity initial) and acceleration (gravity). What equation can you use to solve for T?
    once you get T, u can then multiply it by 2 to get the total time the ball is in the air.

    once u have time, use your x component of velocity to solve for how far the ball travels before it touches the ground. then you can solve for how far the guy has to run to meet the ball, with that distance u have time.
  12. Feb 13, 2009 #11


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    So how long is the ball in the air?

    Figure that out first.

    V = g*t and the total up and down then is t = 2*Vy/g
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