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Two objects free fall

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known data

    a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what
    velocity was the second stone thrown

    2. Relevant equations

    Xf=xi +vt + 0.5t^2

    Vf= vi + at

    3. The attempt at a solution
    The First Stone.

    d1 = -0.5g*t^2 = -4.9*(1.5)^2 = -11 m. =

    Distance traveled by 1st stone after 1,5s.


    Vf= Vi+at.

    Vf= 0 -9.8*1.5 = -14.7m/s

    V = -14.7 m/s = Velocity of 1st stone after 1.5 s.



    d1 = Vo*t + 0.5g*t^2

    -200 = -11 - 14.7 t -4.9t^2

    -4.9t^2 -14.7t +189=0

    Using Quadratic Formula.

    Tf = 4.89 s. = Fall time or time for each stone to reach Gnd.


    The 2nd Stone.

    d = Vo*t + 0.5g*t^2 = -200 m.

    Vo*4.89 -4.9*(4.89)^2 = -200

    Vo= -16.94m/s


    Why Am I getting a negative answer the upward velocity should be positive (the top of the cliff is taking as the X=0 position)
     
  2. jcsd
  3. Mar 31, 2016 #2
    i think you should check your calculation .....
    if the time taken by the 2nd stone to go up and come down is known then it should be equal to 2.V0/g as a body being sent up will come to rest after a time V0/g and fall on ground taking same time.
    taking
     
  4. Mar 31, 2016 #3
    try to think on the event
    1. the 1st. stone is falling through height h so one can calculate time taken say t1 easily using distance traversed.
    2. after 1st. one is dropped the second has been thrown up say after 1.5 sec.
    3. time of flight of second stone can be easily computed as function of initial velocity of throw.
    the above time of flight can be related.
     
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