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Projectile motion with people moving

  1. Oct 29, 2006 #1
    I'm having trouble with this problem as well:

    Quarterback Fred is going to throw a pass to tight end Doug. Doug is 20 m in front of Fred and running straight away at 6.0 m/s when Fred throws the 500 g football at a 40 angle. Doug catches the ball without having to alter his speed and runs for the game-winning touchdown.

    How fast did Fred throw the ball?

    I don't even know where to start with so so if anyone can help me get started that would be great.

  2. jcsd
  3. Oct 29, 2006 #2
    Look at these two equations and figure out how and why they work:



    Solve each in terms of [itex]v_{0}[/itex], equate one equation to the other, solve for t, and then plug it into a formula to find [itex]v_{0}[/itex].
  4. Oct 29, 2006 #3
    Okay, so when i solved for vo i for:

    vo= 6t+20/cost and vo= gt^2/2sint

    then i made these equations equal to each other:

    6t + 20/cost = gt^2/sint

    but when i solve for t, the value that i get donesn't work. if you could explain how to solve for t properly that would be geat.

    Thank you for your response
  5. Oct 29, 2006 #4
    The equation you end up with is a quadratic. Manipulate it to form [tex]ax^{2}+bx+c=0[/tex], and solve for x. You will get two roots[itex]-[/itex]only one will be applicable.
    Last edited: Oct 29, 2006
  6. Oct 29, 2006 #5
    okay this time when i did that i got:

    12sint^2 + 40sint=gcost^3

    -gcost^3+ 12sint^2 + 40sint=0

    t(-gcost^2 + 12sint =40sin)=0

    i solved for inside the brackets using the quadratic formula and i got 2.43 for my time and then i put it back into one of the equations and i got 78m/s however i still got the wrong answer.

    I was wondering if you could please check over my work to make sure that i'm not missing anything. Thanks!
  7. Oct 29, 2006 #6
    2.43 s is the correct time. Which equation did you plug it back into?
  8. Oct 30, 2006 #7
    i put it into the 6t+20/cost equation
  9. Oct 30, 2006 #8
    Put it into:


    Remember, at the time you've specified, y will have a certain value.

    Your question is solved with this equation.
  10. Oct 30, 2006 #9
    okay i got it now, thank you for your help!
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