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Dynamics problem

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    From the edge of a cliff, two stones are thrown at the same time, one vertically upwards and the other vertically downwards with the same velocity of 20 m/s. The second stone reaches the ground in 5 s. How long will the first stone be in air? Also find the height of the cliff.

    2. Relevant equations

    ## s=ut+\frac{1}{2}at^2 ##

    ## v=u+at ##

    3. The attempt at a solution
    I got the height correctly.
    Assuming edge of the cliff as the origin,

    height h=-20*5-0.5*(-9.8)*(5^2)=-222.5 m.

    For the first ball I found two displacements, one while going upwards (s1) and the other downwards (s2).

    s1=20*t1-0.5*9.8*(t12)
    s2=0*t1-0.5*9.8*(t22)=-4.9*(t22)

    since s1=s2
    ∴ 20t1-4.9t12=-4.9t22 .......(1)
    also from v=u+at
    0=20-9.8t1, thus t1=2.04 s
    putting in (1) i got t2=2.04 s.
    The total time the ball was in the air = t1+t2=4.08 s.
    The book gives the answer as 9.08 s. :( Is it wrongly printed? The solution given is:

    222.65=20*t+0.5*9.8*t2. Solving for t, t=9.08 s. But how can the displacement for the first ball be 222.65 m? That's the height of the cliff!!

    Also, is it ok to write displacement s as,
    s(u, t, a) = ut + 0.5*a*t2.
    Will the examiner deduct marks for writing it like this? And u and a are constants so is it just s(t) or s(u, t, a) ?
    Please help. Thanks.
     
    Last edited: Sep 15, 2012
  2. jcsd
  3. Sep 15, 2012 #2
    There's no reason to make two equations for the motion of the ball. The displacement of ball1 can be described using just one equation in the form:

    [itex]\displaystyle s=\frac{at^2}{2}+v_0t+s_0[/itex]

    Since you defined the edge of the cliff as 0, and then you found the time when ball 1's displacement was at 0, you only found the time it took ball1 to be thrown up and then pass the top of the cliff again. You need to find the time it takes to reach the bottom of the cliff.

    In the solution you were given, they defined the positive direction to be downwards, so the final height of ball1 is positive after it drops off the cliff.
     
  4. Sep 15, 2012 #3
    No!! i got the height correctly!
    they plugged in that height as the first ball's displacement which is wrong since it goes up!
    can u also tell me how to write s?
     
  5. Sep 15, 2012 #4
    The first ball starts by going up, but then it falls to the bottom of the cliff.

    I gave you the equation you need for s (displacement), you just need to plug in the values from your problem. In the equation I posted above, a is acceleration (in this case, gravity), v0 is initial velocity, and s0 is initial position (which you defined as 0). Try writing the equation, and if you want to confirm you got it right, you can post it here.

    Once you have that equation, you need to find the time that the ball (ball1) reaches the bottom of the cliff. To do this, you need to figure out where the bottom of the cliff is located on the coordinate system you defined (it sounds like you've done this), and plug in that value for s in your ball1 equation.
     
  6. Sep 15, 2012 #5
    no i mean can u answer my last questions? is it s(u,t,a) or s(t)?
     
    Last edited: Sep 15, 2012
  7. Sep 15, 2012 #6
    Since u and a are constants in this case, it should be s(t). You probably wouldn't get counted off in first year physics unless your professor is strict about notation.
     
  8. Sep 15, 2012 #7
    Okay thanks a lot! :smile: :rofl:
     
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