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Projectile motion, find speed/time of football kick

  1. Sep 4, 2013 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

    A football punter kicks the football at a 30.0 degree angle above the horizontal and just scores a field goal 40.0 meters away, where the horizontal bar is 3.50 meters above the ground.

    a. How long is the ball in the air before the goal is scored?

    b. What was the initial speed of the ball right after the kick?

    This is for college physics. We're doing two dimensional motion/kinematics right now. We just did projectile motion and vectors yesterday.


    3. The attempt at a solution

    My first attempt was to put it in a vector form, using the x movement of 40.0m on the x axis and 3.50 m on the y axis, giving me a vector of (40.0i+3.50j). And I couldn't see where to go from there.

    I tried using the kinematic equation for motion in the y direction-

    [tex]\Delta{x}=v_{0y}t+\frac{1}{2}gt^2[/tex]

    [tex]3.50m=40.0m(\sin30)t+\frac{1}{2}(-9.80)t^2[/tex]

    [tex]4.9t^2-20t+3.50=0[/tex]

    Then I solved the quadratic and I got two solutions of 3.89 seconds, and .183 seconds. The 3.89 seems like reasonable length of time.

    Am I on the right track here, or am I way off?
     
    Last edited: Sep 4, 2013
  2. jcsd
  3. Sep 4, 2013 #2

    QuantumCurt

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    Do I need to find the length of the hypotenuse of the triangle formed by the x and y distances, and use that as the magnitude instead of using the 40.0 m that I multiplied the cos 30 by? That would barely change the value though...it would give me approximately 40.2 m.
     
  4. Sep 4, 2013 #3

    QuantumCurt

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    I'm really feeling like it's not even possible to solve this problem without knowing the velocity or acceleration of the ball.

    The only thing I can see that alludes to velocity is the fact that it "just" made the field goal, which implies that it was the minimum velocity necessary to reach the goal post. Would that be the 40.2 that I get for the length of the hypotenuse of the right triangle?
     
  5. Sep 5, 2013 #4

    gneill

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    You should know that for a projectile trajectory the vertical and horizontal motions are independent. That means you can write an equation for each of them involving time and initial velocity which are your two unknowns. Two equations in two unknowns means you can solve them as simultaneous equations.

    Forget the length of the hypotenuse... The trajectory of the ball is parabolic, not a straight line, and the hypotenuse will not give you any useful information for this type of problem.

    Write out the two kinematic equations for x(t) and y(t) using the given information.

    A couple of points: 1) g is a positive constant. It's better to indicate the effect of the acceleration term by changing the sign of the ##\frac{1}{2}g t^2## terms rather than assuming g to be negative. You will eventually come to problems where you need to consider both rising and falling objects at the same time, and having g be both positive and negative at the same time will be confusing. 2) It's handy to remember a few common trig values. Here you have an angle of 30° and taking its sin and cos, which from geometry you know are ##1/2## and ##\sqrt{3}/2##. This should help to simplify your equations by doing away with the trig expressions.
     
  6. Sep 6, 2013 #5

    QuantumCurt

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    Ok, I think I managed to figure it out.

    Using the kinematic equation for x-direction motion-

    [tex]\Delta{x}=v_{0x}t[/tex]

    [tex]40.0m=v_{0}(\cos 30^o)t[/tex]

    Then solve for t to get-

    [tex]t=\frac{40.0m}{v_{0}(\cos 30^o)}[/tex]


    Then I used the kinematic for y-direction motion-

    [tex]\Delta{y}=v_{0y}t+\frac{1}{2}gt^2[/tex]

    Then substitute my known components in-

    [tex]3.50m=v_{0}(\sin 30^o)t-\frac{1}{2}(9.80 \frac{m}{s^2})t^2[/tex]

    Then substitute the x-direction equation that was solved for t-

    [tex]3.50m=v_{0}(\sin 30^o)(\frac{40.0m}{v_{0}(\cos 30^o)})-\frac{1}{2}(9.80 \frac{m}{s^2})(\frac{40.0m}{v_{0}(\cos 30^o)})^2[/tex]

    Then I get to slog through all of the algebra to solve for [itex]v_{0}[/itex]

    [tex]3.50=\frac{40.0 \ sin 30^o}{\cos 30^o}-4.90(\frac{46.2}{v_{0}})^2[/tex]

    [tex]3.50=23.1-4.90(\frac{2134}{v_{0}^2})[/tex]

    [tex]-19.6=-\frac{10457}{v_{0}^2})[/tex]

    [tex]v_{0}^2=\frac{-10457}{-19.6}[/tex]

    [tex]v_{0}=23.1\frac{m}{s}[/tex]


    Then I substitute my [itex]v_{0}[/itex] back into my equation for t-

    [tex]t=\frac{40.0m}{23.1(\cos 30^o)}[/tex]

    Which then simplifies to [itex]t=2.00 \ sec[/itex]


    Questions- Assuming that I did work through this correctly, which I'm pretty sure I did, is it just by chance that I ended up with a term of 23.1 several steps before I finished solving the equation, or is there some relevance to that? The term that simplified to the 23.1 several steps back did have [itex]v_{0}[/itex] in both the numerator and denominator, but they canceled when I was doing the simplification.

    Would there have been a simpler way of going about this problem? This was the smoothest approach I could find, and it seems to work out pretty well. Any suggestions on things I could have done differently?

    gneill- I took your suggestion of leaving g as a positive value, and just changed my plus sign to a negative sign, rather than substituting a -9.80 into the equation. Is that what you were referring to?

    edit- Another thought. I suppose I would have gotten a more precise answer if I'd left the values of the sines and cosines in exact forms, rather than using decimal approximations. I don't think that would have drastically changed the solution, because I was squaring parts of the equation and taking square roots anyway, but that's definitely something to keep in mind. Due to the need for sig figs, would that really have made much difference here?

    edit 2- Another thought- The algebra in this would have been a lot more straight forward if I'd solved the y-direction kinematic for [itex]v_{0}[/itex] before substituting the actual numbers in. Another important thing to keep in mind for future problems, yes? In this case though, I guess it may not have really helped very much, because I'd still end up having [itex]v_{0}^2[/itex] on both sides of the equation anyway after I substituted my equation for t back in.
     
    Last edited: Sep 6, 2013
  7. Sep 6, 2013 #6

    gneill

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    Yes! Nicely done.

    Looks like a coincidence to me; it boils down to tan(θ)*distance which shouldn't be predictive of speed.
    That's the approach I would take, and you carried it off well. The only thing I might have done differently is to replace the sin and cos with their numerical constants right off the bat since they're trivial constants.
    Yes indeed. Looks good.
    In this case it would not have made a significant difference. That won't always be the case, so using firm constants when available is preferable. When in doubt keep more significant figures. In fact it's a good idea to always carry several extra digits ("guard" digits) through all intermediate calculations to prevent rounding error creeping into the significant figures..
    Actually you might find the algebra easier that way. Solve the y-equation for ##v_o## and substitute it into the x-equation leaving a rather simple expression for ##t^2##.
     
  8. Sep 6, 2013 #7

    QuantumCurt

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    Thanks for the tips!! I'll be sure to remember those in the future. Right after I finished typing this, the fact that I could have substituted exact values in for my sin and cos functions occurred to me. At that point though, I didn't want to work through the whole thing again...lol I did not notice that I could have simplified the sin/cos to a tangent though. I obviously know that sin/cos=tan, just didn't catch it when I was doing the work. That would have made some of the work a little cleaner looking.

    Thanks for the help!!
     
  9. Sep 6, 2013 #8

    gneill

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    You're quite welcome! Cheers!
     
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