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Homework Help: Easy Question

  1. Dec 10, 2004 #1
    A football is kicked at 20m/s from ground level. When the launch angle is 30 degrees, the football will travel a horizontal distance of about_____?

    would this be projectile motion?

    which equation would you use?
  2. jcsd
  3. Dec 10, 2004 #2


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    Yes,u can think of the ball as a projectile.
    HINT:Calculate the intial speed along the 2 axes.Find the time in which the ball is in the air,then use it calculate "x".

    Result:Round about 34m.
  4. Dec 10, 2004 #3
    um...... what's the equation and how do you get the equation
  5. Dec 10, 2004 #4
    How far have you come?

    Here's a few pointers on how you could start to solve the question:

    First, draw the problem on a piece of paper. It's a two-dimensional problem, so the x- and y-axis will do.

    Divide the forces acting on the footboll into the different axis. First, how large is the force acting in the x-axis? Is there anything preventing this force? Secondly, what forces are acting along the y-axis?

    When you have figured out the forces, all you need is to calculate how long it will stay in the air (hint: y=0 will happen twice. When it is kicked, and when it lands).

    Use this time to calculate how far the fotball comes in the x-direction.
  6. Dec 10, 2004 #5
    I suppose I'll just display it in easy terms.

    Find the initial vertical velocity and the initial horizontal velocity.

    horizontal initial velocity = [tex]cos(30) * 20 = 17[/tex]

    vertical initial velocity = [tex]sin(30) * 20 = 10[/tex]

    since 9.8 is the gravitational acceleration, the projectile should reach the top of the path at 10/9.8 seconds, which is ~1. Since the time from the bottom to the top of the parabola is equal to the time from the top to the bottom, we have to multiply this number by 2, which brings us to ~2. Now we know that t = ~2 seconds, and, so we simply multiply this number by the horizontal initial velocity and we should get the total distance (this is brought about by the equation [tex]x = (v_{ix})(\Delta t) + \frac{1}{2}a_xt^2[/tex], but since there is no horizontal acceleration, the second part of the equation drops out, leaving [tex]x = (v_{ix})(\Delta t)[/tex])

    As you can now see, the result is ~34 meters.

    So, as for your question, there is no general equation since there are multiple parts to this problem. Instead, there are 3 or 4 that you have to use to find the final result.
    Last edited: Dec 10, 2004
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