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Two particles are released from the same height at an interval of 1s. How long after the first particle begins to fall will the two particles be 10m apart? 1. 1.5s2. 2s3. 1.25s4. 2.5s

Mahathi , 8 Years ago
Grade 12
anser 3 Answers
Arun
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
Last Activity: 8 Years ago
Shailendra Kumar Sharma
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
Last Activity: 8 Years ago
Gurmeet
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
Last Activity: 8 Years ago
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