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I think this question is impossible.

  1. Jan 1, 2006 #1
    a football is kicked at an angle of 45 degrees and it travels 82m before it hits the ground. Ignore air resistance

    a. What is its initial velocity?

    b. How long was it in the air?

    c. How high did it go?
  2. jcsd
  3. Jan 1, 2006 #2


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    Well, the only force acting on the ball is gravity. This means you can write parametric equations for the ball's position:
    (assuming the ball starts at (0,0))
    When the ball hits the ground, y=0. Using this you can start answering the questions.
  4. Jan 1, 2006 #3
    .......? that confused me even more. You don't know the time, so I dont see how you can do it.
  5. Jan 1, 2006 #4
    Have you done an energy unit yet?
  6. Jan 1, 2006 #5
    no i have not good sir/mam
  7. Jan 1, 2006 #6
    ok, for a i got 20
    b. 4.1
  8. Jan 1, 2006 #7


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    Exactly, you don't know the time: that's what you're solving for! You have
    So when the ball hits the ground, y=0. This means:
    You know that the ball starts off on the ground at t=0, you want to find the second time the ball hits the ground when t is not zero. Since t is not 0 you can divide it out to get:
    Since the ball was kicked at a 45° angle v_0x=v_0y, so
    The second piece of information you're given is that the ball is 82m away at this time, or x=82m. Substitute in for time in the x equation and solve for v_0x. Once you've done this you can plug this into the equation for time to get a numeric answer for the time elapsed. Keep in mind that the total velocity is a vector when answering the first part.
  9. Jan 1, 2006 #8


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    HINT: Textbook authors do not make a habit of putting impossible questions into the exercise sections. :)
  10. Jan 1, 2006 #9


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    Keep in mind that the velocity is a vector; it has two components. This is only one component of the velocity.
    Almost right. Units?
    The path of the ball is a parabola. to see this, solve for time in the x equation:
    [tex]\rightarrow t=\frac{x}{v_{0x}}[/tex]
    Then sub this into the y equation:
    [tex]\rightarrow y=v_{0y}\frac{x}{v_{0x}}-\frac{1}{2}g(\frac{x}{v_{0x}})^2 [/tex]
    [tex]y=x-\frac{1}{2}g(\frac{x}{v_{0x}})^2[/tex] (remember [tex]v_{0x}=v_{0y}[/tex])
    Recall that the formula for the turning point of a parabola with equation
    [tex]y=ax^2 + bx +c[/tex]
    Last edited: Jan 1, 2006
  11. Jan 2, 2006 #10
    The standard equation for the range is

    [tex]R = \frac{|V|^{2}\sin 2\theta}{g}[/tex]

    You should have the derivation of this in your notes, or you can use the method Euler has posted above multiplying the time in the air with the horizontal speed. Since you have |V| and the direction, you have the velocity.

    For the 3rd part, you have worked out the time it is in the air for the 2nd part, so just apply that to purely vertical motion, where the initial speed is the vertical speed of the particle, or use Euler's method which is to give the height of the motion in terms of the distance travelled.
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