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Some Issues with Projectile Motion

  1. Sep 19, 2004 #1
    Hello again,

    I'm having some issues with two of my problems. But I'll do one at a time and keep them here without having to start mulitple threads clogging the forum.

    Here's my first problem: A soccer player kicks the ball toward a goal that is 28.0m in front of him. The ball leaves his foot at a speed of 19.1m/s and an angle of 33.8 degrees above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is NOT 19.1 m/s.)

    So here is my reasoning and what I have done. I assumed that if I found the accleration of the ball in the x direction and then pluged that answer back into a velocity equation, I would have the answer. But of course, I do not.

    I also found Voy and Vox. I found Voy using: Voy = Vo sin theta and then Vox = Vo cos theta. From there I thought I would need to find the time the ball was in the air for the y axis, but now I'm thinking perhaps I should have found how long the ball was in the air along the x axis. But if that were the case I wouldn't know what the acceleration was.

    In any case from everything I've done, I don't have the correct answer. Am I making this harder than it really is? I have a terrible problem with that. Was I at least even headed in the right direction? Thanks for any help provided.
  2. jcsd
  3. Sep 19, 2004 #2


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    Homework Helper

    On x-axis there's constant velocity.
  4. Sep 19, 2004 #3
    Heres a general formula for these type of questions.

    First resolve the components of the velocity vertically and horizontally.

    You know (assuming no air resistance) that the horizontal component of the ball is not subject to any force so it will remain constant. You know the horizontal distance travelled by the ball... 28m so you can find the time of flight for the ball from this "horizontal" information.

    Now you know the inital vertical velocity of the ball. You know the force on the ball (its just gravity) and you know the time of flight for the ball. So from this you can work out the vertical velocity of the ball after it has travelled the 28m horizontally.

    Now you have the horizontal component of velocity (it's the same as the inital horizontal velocity) and you have just calculated the vertical component of velocity. So from these you can work out the speed of the ball.
  5. Sep 19, 2004 #4

    Doc Al

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    Staff: Mentor

    The only acceleration is that due to gravity, which acts only in the y-direction (vertical).

    You know the horizontal distance and speed (the x-component of the velocity stays constant). Use that to find the time.

    Then find the y-component of the velocity at that time.
  6. Sep 19, 2004 #5
    Ok, I think I understood. So here's what I did, but my answer doesn't seem right. To me anyway.

    Since velocity is constant Vox and V is the same. For Vox my answer was 15.9 (I rounded to save writing a rediculously long number.) So V is the same since velocity is constant. I used this equation: X = 1/2 (Vox + V)t to find time. For which I got 1.76. I then used this equation V = Vo + at to find the final speed, however my answer was 1.81. Which doesn't seem right to me.

    What am I not understanding here?
  7. Sep 19, 2004 #6
    Never mind. (1) I realized my last equation was wrong. I should have used : Vy = Voy + at. And of course a problem is never finished without using Pythagorean's thereom. Silly me. So I've got this one situated. I'm gonna give the other one a go, one more time by myself to see if I can get it.
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