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Confused kinematics equations & ball motion in the air

  1. Sep 16, 2012 #1
    you throw a ball straight up in the air with a speed of 7.25m/s. The moment the ball leaves your hand you start running away at a speed of 2.59m/s. How far are you from the ball the moment it hits the ground.

    So I've gotten better at these kinematics equations but this one has too many factors..I don't even know where to begin. I was thinking that I could use y=y0+ v0t +1/2at2 but what would I solve for?? Then would I put it into the same equation but for x???
     
  2. jcsd
  3. Sep 16, 2012 #2

    lewando

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    Re: Confused... please help

    Solve for t. Since you are not accelerating when you are running, you could-- it reduces to d=r*t.
     
  4. Sep 16, 2012 #3
    I'm still confused...if I try to solve for t what is my change in y? It says 4ft from the ground....but what about what it travels when it's is thrown up? Also wouldn't I have to make a= 9.81?
     
  5. Sep 16, 2012 #4

    lewando

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    4ft is new information. I assume that would be your y(0). You are solving for T when y(T)= 0. Yes, a is 9.81m/s^2 but in what direction?
     
  6. Sep 16, 2012 #5
    Oops yes it is thrown from a height of 4ft. So Would I still make my y and y0 = 0 and them handle the 4 ft afterwards?
     
  7. Sep 16, 2012 #6

    HallsofIvy

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    A ball going upward with initial speed 7.25 m/s and initial height 4m will have height h= -4.9t^2+ 7.25t+ 4 at time t. When the ball "hits the ground", h= 0, of course so you want to solve -4.9t^2+ 7.25t+ 4= 0. That's a quadratic equation and will have two solutions but one will be negative. (It will be the time the ball would have been on the ground if it were thrown upward from the ground to be at height 4m with speed 7.25 at time t= 0.)
     
    Last edited: Sep 16, 2012
  8. Sep 16, 2012 #7

    lewando

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    No-- y = 0 (your y of interest when the ball hits the ground) and y0 = 4 (initial height).
     
  9. Sep 16, 2012 #8
    So the quadratic eqtn would give me the time it would take .... So I would then have to plug this into the eqtn x=x0+v0t+1/2at^2 but it would become x=x0+v0t since a for x equals zero... Right?
     
  10. Sep 16, 2012 #9

    lewando

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    Right. Watch your sign on the ay term, show us your full equations.
     
  11. Sep 16, 2012 #10
    Hmmm... I'm still getting it wrong. When I put into the quadratic it's 1.21m + 7.25 t + -4.9t2. Btw it's 1.21m from the 4ft given... I get a time that is a decimal so when I put it into the x eqtn it's very low.... I'm stuck
     
  12. Sep 16, 2012 #11

    lewando

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    Show us your equations that you are about to evaluate. You are possibly very close.[edit-- what did you get for T?, for x?]
     
    Last edited: Sep 16, 2012
  13. Sep 16, 2012 #12
    Ok, the #'s change each time but same question so, given y0=4ft(1.21m) v0y=5.16m/s v0x=2.41.

    Quad eqtn 1.21+5.16t+(-4.91)t^2

    End up with -5.16+-sqrt50.3/2(-4.9) T=1.25. So when I plug this into the x eqtn x=v0t. X=2.41•1.25.
    X= 3.01m

    I dont know I this is right, once I enter it the #'s will switch again... Does it look right?
     
  14. Sep 16, 2012 #13

    lewando

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    It is getting late. Would expect to see formula for the y activity to be this:

    0 = 1.21 +7.25*t -0.5*9.81*t2

    where did you get 5.15m/s2?

    not sure I follow--"the numbers change all the time"
     
  15. Sep 16, 2012 #14
    Yes this is what I had, but I mentioned the #'s changed when i entered my answer so 7.25 became 5.16. It is late, thank you for your help... I wouldn't have gotten this far so thank u and good night :)
     
  16. Sep 16, 2012 #15

    lewando

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    Continue the thread when you can. Many awesome folks are here who can help move it along.
     
  17. Sep 16, 2012 #16
    Yay, actually I got it! Thanks sobmuch!
     
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