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A particle A dropped from rest at a height h above the ground at time t=0, falls...
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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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)`
`h = uT-(gT^2)/2+(gT^2)/2`
`T = h/u`
So the two particles meet each other when the time is `h/u` .
Posted by jeew-m on September 23, 2013 at 8:28 PM (Answer #1)
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