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Robinronaldo

  • 2 years ago

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?

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  1. exploringphysics
    • 2 years ago
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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

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

  3. shamim
    • 2 years ago
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    ur answer will b 148.48 m

  4. shamim
    • 2 years ago
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    can i explain my work

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

  6. shamim
    • 2 years ago
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    no

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

  8. shamim
    • 2 years ago
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    for full height u=0

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

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

  11. shamim
    • 2 years ago
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    its my 1st equation

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

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

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

  15. agent0smith
    • one year ago
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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. \[h = \frac{ 1 }{ 2} g t ^{2}\] \[\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t - 1) ^{2}\]

  16. chongkhengwye
    • one year ago
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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

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