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anonymous

  • 5 years ago

2 stones are dropped from the top of a tower at half a second apart .The time after dropping the first stone at which the distance between the 2 stones is 20 m is a. 4 s b. 4.25 s i took 1 distance as x and other as 20 + x ... how to take time ..can some1 pls help me...

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  1. anonymous
    • 5 years ago
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    Since intially the time difference was 0.5s, take the time taken by the first stone (till the instant when distance is 20) as t, then other stone takes t-0.5. You could do this way or an easy method is to use the concept of the relative velocity. Using that when the second stone is dropped, the first stone has a velocity of 0.5*10=5 and is (0.5)*(10)*(0.5)^2=1.25m below first stone. After dropping the second stone, when you see the relative motion of first stone wrt the second stone, it is uniform motion without any relative acceleration (As both are in free-fall) and the time taken for the distance to be 20m, the second stone should gain 18.75m more from the dropping of second stone which takes 18.75/5=3.75s. So the time from dropping of first stone turns out to be 3.75+0.5=4.25s.

  2. anonymous
    • 5 years ago
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    tnx a lot.. :) u're gr8 !!

  3. anonymous
    • 5 years ago
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    can u pls help once more?? i dint understand 18.75/5..y did u do that?

  4. anonymous
    • 5 years ago
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    is that distance/velocity =time?

  5. anonymous
    • 5 years ago
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    Yeah, it is. I could do that since it is uniform velocity motion.

  6. anonymous
    • 5 years ago
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    thank u...!

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