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Robinronaldo

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

  • one year ago
  • one year ago

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  1. exploringphysics
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    distance covered inlast sec. is v+1/2g=v+5=1/3h; v=([2h/g]^1/2 -1)g; v^2=(h/5+1-2[h/5]^1/2)g^2; solve these and eliminatte v you get h

    • one year ago
  2. Robinronaldo
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    @ashwinjohn3 what would be the answer in your opinion?

    • one year ago
  3. shamim
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    ur answer will b 148.48 m

    • one year ago
  4. shamim
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    can i explain my work

    • one year ago
  5. ashwinjohn3
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    ok use the equation....... Diplacement @ nth second=u+ a(2n-1)/2

    • one year ago
  6. shamim
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    no

    • one year ago
  7. shamim
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    i wanna use the equation\[h=ut+ \frac{ 1 }{ 2 } g t ^{2}\]

    • one year ago
  8. shamim
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    for full height u=0

    • one year ago
  9. shamim
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    and my equation become \[h=0+\frac{ 1 }{ 2 } g t ^{2}\]

    • one year ago
  10. shamim
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    \[h=\frac{ 1 }{ 2 } \times 10 \times t ^{2}=5 \times t ^{2}\]

    • one year ago
  11. shamim
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    its my 1st equation

    • one year ago
  12. shamim
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    now i wanna write another equation for 1st 2/3 h height

    • one year ago
  13. shamim
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    \[\frac{ 2 }{ 3 }h=ut+\frac{ 1 }{ 2 } \times g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times (t-1)^{2}\]

    • one year ago
  14. shamim
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    now u hv to solve these 2 equation to get time t and then u will get the height h

    • one year ago
  15. agent0smith
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    You will get two values for t, but one of them will be less than 1 second, making it invalid, as you can see from the question "covers one-third of its total distance to the ground in the *****last second***** of its fall" Solve the two equations shamim gave you, by substituting the first one into the second, then solve for t. The second equation comes from the fact that, if it covers one third of the total distance in the last second, then it must cover the initial 2/3rds of the total distance in (t-1) seconds. \[h = \frac{ 1 }{ 2} g t ^{2}\] \[\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t - 1) ^{2}\]

    • one year ago
  16. chongkhengwye
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    Say the height of the cliff is H, time take for fall is T, and g is the acceleration of freefall. \[H=\frac{ 1 }{ 2 }gT ^{2}\] --- (1) \[\frac{ 2 }{3} H=\frac{ 1 }{ 2 }g(T -1)^{2}\] --- (2) Solving you will get 2 answer for T: 0.551 s and 5.45 s But T>1, Hence H = 146 m

    • one year ago
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