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Grade: 12

                        

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

3 years ago

Answers : (3)

Arun
24736 Points
							
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
2 years ago
Shailendra Kumar Sharma
188 Points
							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
						
2 years ago
Gurmeet
46 Points
							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
						
2 years ago
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