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#. The particles collide when the top one has fallen a distance #y#.
The height of point A above point B is #h#.
If #h=100m#, #v=20 ms^-1# and #g=10 ms^-2#, sketch the velocity-time graph, and mark on the graph the #(t,v)# values where the particles collide. Take #v# to be positive upwards.

An earlier part of the question also had you showing that the time to collision was #t=h/v#

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Answer 1 · 2018-02-24T16:21:51

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

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

# y=1/2g t^2#...1

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

#100-y = 20t - 1/2g t^2#

or, #100 -1/2g t^2 = 20 t -1/2 g t^2#(from 1)

so, #t=100/20 = h/v=5 s# (given, #h=100# and #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)