# A particle A dropped from rest at a height h above the ground at time t=0, falls vertically under gravity. At the same instant another particle B is projected vertically upwards from a point on the...

A particle A dropped from rest at a height h above the ground at time t=0, falls vertically under gravity. At the same instant another particle B is projected vertically upwards from a point on the ground with velocity u. Draw the velocity-time graph for the motion of each particle on the same diagram.

Using the velocity-time graphs, show that the two particles are at the same height from the ground at time h/u.

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### 1 Answer

See attached image with the explanations.

The red graph shows the particle A that drops from height h and green graph shows that of the projected particle B.

Both have the same acceleration due to gravity but at different directions.

Particle A is accelerating and particle B is decelerating.

Let us say that the particles is at same height at time t = T.

If the particle B travels distance H in this time then particle A has traveled h-H in the time interval.

According to the graph;

`v = gT`

`(u-w)/T = -g rarr w = u-gT`

Using the area of the graphs;

For particle B

`H = (u+w)/2xxT = (2u-gT)/2xxT---(1)`

For paricle A

`h-H = (vT)/2 = (gT^2)/2 ----(2)`

(1)+(2)

`h = uT-(gT^2)/2+(gT^2)/2`

`T = h/u`

*So the two particles meet each other when the time is `h/u` .*

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