Let the objects be O1 and O2,when the given situation appears, Let S1=x , therefore S2=x-10. Let t2=t, since O1 started 1s earlier,therefore t1=t+1. Now since S=ut+at^2/2 => S=at^2/2 (as u=o).➡eq1: x=9.8×(t+1)^2 /2 & eq2: x-10=9.8×t2 /2 . Now eq1-eq2 => +10=9.8(t2+1+2t)/2 - 9.8×t2 /2 => 10=9.8(t2+1+2t-t2) /2 => 20=9.8+19.6t => 10.2/19.6=t => t=0.5(approx.) therefore time take for the process= t+1 =1+0.5=1.5s
						

							Distance travelled in 1 sec by first particle Is 1/2(gt2)= 1/2(9.8*1)= 4.9 mVelocity at this time = at= 9.8Now these 2 particles moves with a relative velocity of 9.8 m/s So Time taken to cover remaining 5.1 m will be 5.1/9.8=.52 So total time will be 1.52 sec
						

							Let the objects be O1 and O2,when the given situation appears, Let S1=x , therefore S2=x-10. Let t2=t, since O1 started 1s earlier,therefore t1=t+1. Now since S=ut+at^2/2 => S=at^2/2 (as u=o).➡eq1: x=9.8×(t+1)^2 /2 & eq2: x-10=9.8×t2 /2 . Now eq1-eq2 => +10=9.8(t2+1+2t)/2 - 9.8×t2 /2 => 10=9.8(t2+1+2t-t2) /2 => 20=9.8+19.6t => 10.2/19.6=t => t=0.5(approx.) therefore time take for the process= t+1 =1+0.5=1.5s17 day