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A stone is dropped from height h. Simulataneously another stone is thrown up from ground with such a velocity that it can reach a height of 4h. Find the time when two stones cross each other.
the second ball can reach a height of 4h. so, at 4h its velocity =0 v2=u2+2as so, 0 =u2 -2g*4h so, u=sqrt(80h)i have taken upward direction to be positive.for the dropped stone, u=0 and for the thrown stone, u=sqrt(80h)let they cross at time t and the dropped ball covers distance/displacement x , then the thrown ball covers h-x.so, for the dropped ball, -x=0-(1/2)g*t2 and the second ball, h-x = [sqrt(80h)]*t -(1/2)gt2 sobtracting the second equation from the first one, we get h=[sqrt(80h)]*t so, t=sqrt(h/80)
v2=root(8gh)v1=root(2gh)relative v=v1+v2t=h/(v1+v2)
Thank you sir for this answer. Thank you very much.
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