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

## 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 $h = 100 m$, $v = 20 m {s}^{-} 1$ and $g = 10 m {s}^{-} 2$, sketch the velocity-time graph, and mark on the graph the $\left(t , v\right)$ 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 $t = \frac{h}{v}$

Feb 24, 2018

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,

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

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

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

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

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

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

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

So,when both will meet,the velocity of th one going down will be $- 10 \cdot 5 = 50 \frac{m}{s}$ and the one going up will be $20 - 10 \cdot 5 = 30 \frac{m}{s}$

Now,these things are plotted to make a graph

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 $- 10 \cdot 2 = - 20 \frac{m}{s}$ (downwards)