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Can't remember how I got these answers

  1. Sep 20, 2007 #1
    The kicker on a football team can give the ball an initial speed of 26.2 m/s. If he is to score a field goal from a point 42.2 m in front of goalposts whose horizontal bar is 3.75 m above the ground, what is the minimum angle above the horizontal he must kick the ball?

    I was suppose to calculate the minimum and maximum angles.

    I got 24.3 and 70.8.

    I set up this equation:
    xf=xi+vi*cos(theta)t
    yf=yi+vi*sin(theta)t+4.9t^2

    if you solve x time, and then sub it in, it comes out to

    yf=yi+tan(theta)x+(4.9x^2/v^2)tan^2(theta)

    3.75=0+tan(theta)42.2+(4.9(42.2)^2/26.2^2)tan^2(theta)

    I'm not sure where I'm making the error.

    I think I did this right, we were told to use the quadratic formula and treat the tan as variables. I'm not really sure what I did after that however.
     
  2. jcsd
  3. Sep 20, 2007 #2

    You have made a mistake in deriving the final formula. Try deriving the formula again.
     
  4. Sep 20, 2007 #3
    I re-derived it and got the same thing.
     
  5. Sep 20, 2007 #4
    I'm not really finding an error, it must be something I missed at the very beginning.

    Would you be referring to the 4.9's position within the parantheses. I didn't have it like that in my calculations.
     
  6. Sep 20, 2007 #5
    Ok let me tell you that the equation is

    y = [tex]\tan\theta[/tex]x - [tex]\frac{gx^2}{2v^2\cos^2\theta}[/tex]
     
  7. Sep 20, 2007 #6
    yes, but doesn't 1/cos^2=1+tan^2

    ahhh i forgot to sub in 1+tan^2
     
  8. Sep 20, 2007 #7
    Of course it is.But you didnt substitute it. Ok,so got the solution.
     
  9. Sep 20, 2007 #8
    Since the hint was to plug that back into the quadratic formula. I'm just not sure what the C value would be.
     
  10. Sep 20, 2007 #9
    I'm still getting an answer that's way off
     
  11. Sep 21, 2007 #10
    Cmon its taking too long.Its now just a quadratic equation.You can solve it now with much ease.But if you are not able to get then i am here.

    I have got the answer but i cannot give it to you as it will break the forum rules.
     
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