MHB Proving |PR|=3|RQ|: Two Particles Collide

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The discussion focuses on proving the relationship |PR|=3|RQ| in a scenario where two particles collide after being projected and dropped from points P and Q, respectively. The first particle is projected upwards while the second falls from rest, and they collide at point R after time t. Using the equations of motion, the position of each particle is analyzed, leading to the conclusion that their speeds are equal at the moment of collision. The calculations show that the distance from P to R is three times the distance from R to Q, confirming the relationship |PR|=3|RQ|. This mathematical proof effectively demonstrates the required relationship through kinematic equations.
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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 traveling 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 don't know where to go from here
 
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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:

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

And the second particle's position is:

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

Now, when the two particles meet, we have:

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

At this time, their speeds are equal, hence:

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

Hence:

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

And so we find:

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

And this implies:

$$\overline{pr}=3\overline{rq}$$
 
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