anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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
anonymous
  • anonymous
@ashwinjohn3 what would be the answer in your opinion?
shamim
  • shamim
ur answer will b 148.48 m

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shamim
  • shamim
can i explain my work
anonymous
  • anonymous
ok use the equation....... Diplacement @ nth second=u+ a(2n-1)/2
shamim
  • shamim
no
shamim
  • shamim
i wanna use the equation\[h=ut+ \frac{ 1 }{ 2 } g t ^{2}\]
shamim
  • shamim
for full height u=0
shamim
  • shamim
and my equation become \[h=0+\frac{ 1 }{ 2 } g t ^{2}\]
shamim
  • shamim
\[h=\frac{ 1 }{ 2 } \times 10 \times t ^{2}=5 \times t ^{2}\]
shamim
  • shamim
its my 1st equation
shamim
  • shamim
now i wanna write another equation for 1st 2/3 h height
shamim
  • shamim
\[\frac{ 2 }{ 3 }h=ut+\frac{ 1 }{ 2 } \times g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times (t-1)^{2}\]
shamim
  • shamim
now u hv to solve these 2 equation to get time t and then u will get the height h
agent0smith
  • agent0smith
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. \[h = \frac{ 1 }{ 2} g t ^{2}\] \[\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t - 1) ^{2}\]
anonymous
  • anonymous
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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