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  1. MHB Craftsman

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    #1
    a particle is projected vertically upwards from a point P. at the same instant a second particle is let fall from rest vertically at q. q is directly above p. the 2 particles collide at a point r after t seconds . when the 2 particles collide they are travelling at equal speeds. prove that |pr|=3|rq|

    i am trying to solve this with uvast equations for the first particle i have v=v s=r a=-g t=t for the second particle i have v=v s=p-r a=g t=t dont know where to go from here

  2. Pessimist Singularitarian
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    #2
    I would begin by orienting the vertical axis of motion such that point $p$ is at the origin. The first particle's position is then given by:

    $ \displaystyle x_1=-\frac{a}{2}t^2+v_0t$

    And the second particle's position is:

    $ \displaystyle x_2=-\frac{a}{2}t^2+q$

    Now, when the two particles meet, we have:

    $ \displaystyle x_1=x_2\implies t=\frac{q}{v_0}$

    At this time, their speeds are equal, hence:

    $ \displaystyle -at+v_0=at\implies v_0=2at$

    Hence:

    $ \displaystyle t^2=\frac{q}{2a}$

    And so we find:

    $ \displaystyle r=-\frac{a}{2}\cdot\frac{q}{2a}+q=\frac{3}{4}q$

    And this implies:

    $ \displaystyle \overline{pr}=3\overline{rq}$

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