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Motion of a Soccer Ball

  1. Sep 14, 2013 #1
    Neglect air resistance for the following. A soccer ball is kicked from the ground into the air. When the ball is at a height of 10.07 m, its velocity is (vx,vy) = (4.63,3.83) m/s.
    To what maximum height will the ball rise?
    What horizontal distance will be traveled by the ball?
    With what velocity (magnitude and direction) will it hit the ground?

    I haven't started the second and third question yet. I wanted to use the equation y=1/2gt^2+Viy*t but I don't know what time or y is. Then I tried y=yo+tan(θ)x+g/(2v^2*cos(θ)^2)*x^2, but I don't have enough info to use it. Would I be wrong to move 10.07 to the origin, then use 0=1/2gt^2 to find maximum height?
     
  2. jcsd
  3. Sep 14, 2013 #2
    What the ball reaches its max height, what is its vertical velocity? How much time will that take from any known value of the vertical velocity?
     
  4. Sep 15, 2013 #3
    dory, I was braindead last night. I figured out that the max height is 10.82 m and the range is 13.75 m. although the last one is still bothering me.
    I found the time it took the ball to hit the ground from height=10.82 which was about 1.47 s. I used 1.47 in Vy=gt (because Vy at max height is 0) and came up with -7.355 m/s. so I used Pythagorean theorm to find the velocity when it hits the ground. I came up with 8.69 m/s and used atan(-7.355/4.63) to get -57.81 deg. But the site says its wrong. I don't know which number Is wrong or if I did a step wrong.
     
  5. Sep 15, 2013 #4
    sorry*
    guess I still am -_-
     
  6. Sep 15, 2013 #5
    The max height and range you got are correct. The time from the max height to the ground is correct, too. However, the final vertical velocity is wrong. Calculate it again from the formula you have.
     
  7. Sep 15, 2013 #6
    ohh.. I have bad handwritind -_- thank you. I did get the right answers
     
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