Sketch the velocity-time graph for two particles colliding in vertical motion?

Question

Two points A and B are located on a vertical line with point A vertically above point B. A particle is released from rest at A and at the same time another particle is projected vertically upwards from B at velocity $v$ $v$ . The particles collide when the top one has fallen a distance $y$ $y$ .
The height of point A above point B is $h$ $h$ .
If $h = 100 m$ $h=100m$ , $v = 20 m s^{- 1}$ $v=20 ms^-1$ and $g = 10 m s^{- 2}$ $g=10 ms^-2$ , sketch the velocity-time graph, and mark on the graph the $(t, v)$ $(t,v)$ values where the particles collide. Take $v$ $v$ to be positive upwards.

An earlier part of the question also had you showing that the time to collision was $t = \frac{h}{v}$ $t=h/v$

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Answer 1 · 2018-02-24T16:21:51

Let,after time $t$ $t$ there will be collision between the two particles,

So,for the one,coming down in that time if comes down by distance $x$ $x$ ,we can write,

$y = \frac{1}{2} g t^{2}$ $y=1/2g t^2$ ...1

so, for the one going up,we can write,

$100 - y = 20 t - \frac{1}{2} g t^{2}$ $100-y = 20t - 1/2g t^2$

or, $100 - \frac{1}{2} g t^{2} = 20 t - \frac{1}{2} g t^{2}$ $100 -1/2g t^2 = 20 t -1/2 g t^2$ (from 1)

so, $t = \frac{100}{20} = \frac{h}{v} = 5 s$ $t=100/20 = h/v=5 s$ (given, $h = 100$ $h=100$ and $v = 20$ $v=20$ )

Now,equation of velocity time relationship for the particle going up is $v'=v-g t=20 -10t$ (taking upward positive)

and, for the one going down is $v=-g t=-10t$

So,when both will meet,the velocity of th one going down will be $-10*5=50 m/s$ and the one going up will be $20-10*5=30 m/s$

Now,these things are plotted to make a graph enter image source here

Sorry for the drawing got excessively enlarged,but I have tried to mark velocities of the two particles for important time values.

Note, at $t=2$ the particle going up comes to rest,but the one going down has a velocity of $-10*2=-20 m/s$ (downwards)