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Two particles are released from the same height at an interval of 1s. How long after the first particle beings to fall will the two particles be 10 m apart?
For Particle A (say) it must have travelled some distance in 1 sec after the second particle has been released.at 1 sec,finalveloccityv = 0 + g*1 = 10 m./sFor particle B,after 1 s when it fell,x2 = 0 + 0.5*g*t^2..............(1)Now,Agin writing their distnace eqns after this 1 sec.y1 = 10*t + 0.5*g*t^2............(2)given x2 – y1 = 10 m.................(3)solving the eqns. (1) (2) and (3) get t.
Let the ball which is thrown first be AAnd the ball thrown after 1sec. be BLet time taken by ball A be t sec.And by ball be (t-1) sec.Let ball A reach at x than ball B will reach at (x-10) For ball A:S=-x ,u=0 ,time=t ,acc.=-10S=ut+1/2at^2-X= 0(t)-5t^2X=5t^2 (let this be eq. No. 1)For ball B:S=-(x-10) , u=0 , time=t-1, acc.=-10S=ut+1/2at^2-(x-10)= -5(t-1)^2 (since u=0)Putting value from eq. 1:5t^2-10=5t^2+5-10tT=15/10=1.5 sec. Put value in eq. 1and find x=11.25 m
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