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    A particle is projected vertically upwards at 20ms from point P.1s later a second particle is projected vertically upwards at same velocity from P . How long after second particle was projected do the 2 particles collide
    Answer is 1.54 from start of second one
    How to get 1.54
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    (Original post by Hasham123)
    A particle is projected vertically upwards at 20ms from point P.1s later a second particle is projected vertically upwards at same velocity from P . How long after second particle was projected do the 2 particles collide
    Answer is 1.54 from start of second one
    How to get 1.54
    The balls collide when they have the same displacement from the starting point.

    You can use one of the suvat equations (which includes s, u, t, and a) on each of the particles. If particle 1 is at time t in its flight, then particle 2 will be at time t-1 in its flight since it was launched one second later.

    Set the displacements equal to each other and solve for t.

    Please post your workings
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    So do s= uvxt +1/2at^2 assign s a value say 5 then do
    5=uvxt +1/2at^2
    5=uvx(t-1) +1/2ax(t-1)2
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    (Original post by Hasham123)
    So do s= uvxt +1/2at^2 assign s a value say 5 then do
    5=uvxt +1/2at^2
    5=uvx(t-1) +1/2ax(t-1)2
    That's basically the right method, but check your SUVAT equations... should be s=ut+\frac{1}{2}at^2, where u=20m/s and a = g = -9.81m/s^2.

    Why are you setting the value of s=5?

    You have two equations with s=..., therefore both equations equal each other.

    Hence you can solve for t.
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    ut + 0.5*-9.8*t2 = u(t-1) + 0.5*-9.8*(t - 1)2

    no need to put in a number such as 5.

    put u = 20 m/s into both sides.
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    So S=20t -4.9t^2
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    Wait is it simaltanius equations
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    (Original post by Hasham123)
    Wait is it simaltanius equations
    Yes, you have two equations with s=... therefore you can solve them simultaneously by setting them equal to each other so there are just ts.
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    So
    S=20t-4.9t^2
    S=20(t-1) -4.9(t-1)^2
    Multiply out the brackets then solve for t?
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    (Original post by Hasham123)
    So
    S=20t-4.9t^2
    S=20(t-1) -4.9(t-1)^2
    Multiply out the brackets then solve for t?
    Yes, so 20t-4.905t^2 = 20(t-1)-4.905(t-1)^2

    And solve as you said.

    Note that in AS physics, most boards ask for g=9.81, whereas maths exams tend to use g=9.8

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    Amateur question but do I times the bracket by outside number before squaring it or square it then times the answer by outside number
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    (Original post by Hasham123)
    Amateur question but do I times the bracket by outside number before squaring it or square it then times the answer by outside number
    Multiply out the brackets using FOIL, then multiply each term by the constant.
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    So do (t-1)^2 then times answer by -4.905
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    (Original post by Hasham123)
    So do (t-1)^2 then times answer by -4.905
    Yes, but you need to expand (t-1)^2 using FOIL:

    Spoiler:
    Show

    -4.905(t-1)^2 \equiv -4.905(t^2 - 2t + 1) \equiv -4.905t^2 + 9.81t - 4.905
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    Foil?
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    (Original post by Hasham123)
    Foil?
    First, Outer, Inner, Last

    It's a method for expanding brackets that you should have met at GCSE.

    Note that (t-1)^2 = (t-1)(t-1), so there you have a pair of brackets to expand.
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    24.905=-49.5t
    T=0.5031 which is not right
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    Whoops that's wrong messed up
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    24.905=9.8t
    T=2.54
    Then I just take away 1 then I'm at 1.54
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    (Original post by Hasham123)
    Whoops that's wrong messed up
    I need to go soon, but I'll show my working in spoiler in case you still need it:

    Spoiler:
    Show

    s= ut-\frac{1}{2}gt^2 = u(t-1) - \frac{1}{2}g(t-1)^2

    \Rightarrow ut -\frac{1}{2}gt^2 = ut - u - \frac{1}{2}gt^2 + gt - \frac{1}{2}g

    \Rightarrow gt = \frac{1}{2}g + u

    \Rightarrow t = \frac{1}{2} + \frac{u}{g}

    \Rightarrow t = 0.5 + \dfrac{20}{9.81} = 2.54 to 3sf

    Now we defined t as the time after the first ball was launched. The second ball was released one second later, therefore they collide 1.54s after the second ball was launched.

 
 
 
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