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 2 years ago
A rock dropped from a cliff covers onethird of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?
 2 years ago
A rock dropped from a cliff covers onethird 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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exploringphysics
 2 years ago
Best ResponseYou've already chosen the best response.0distance covered inlast sec. is v+1/2g=v+5=1/3h; v=([2h/g]^1/2 1)g; v^2=(h/5+12[h/5]^1/2)g^2; solve these and eliminatte v you get h

Robinronaldo
 2 years ago
Best ResponseYou've already chosen the best response.0@ashwinjohn3 what would be the answer in your opinion?

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1ur answer will b 148.48 m

ashwinjohn3
 2 years ago
Best ResponseYou've already chosen the best response.0ok use the equation....... Diplacement @ nth second=u+ a(2n1)/2

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1i wanna use the equation\[h=ut+ \frac{ 1 }{ 2 } g t ^{2}\]

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1and my equation become \[h=0+\frac{ 1 }{ 2 } g t ^{2}\]

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1\[h=\frac{ 1 }{ 2 } \times 10 \times t ^{2}=5 \times t ^{2}\]

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1now i wanna write another equation for 1st 2/3 h height

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ 2 }{ 3 }h=ut+\frac{ 1 }{ 2 } \times g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times (t1)^{2}\]

shamim
 2 years ago
Best ResponseYou've already chosen the best response.1now u hv to solve these 2 equation to get time t and then u will get the height h

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.0You 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 onethird 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 (t1) seconds. \[h = \frac{ 1 }{ 2} g t ^{2}\] \[\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t  1) ^{2}\]

chongkhengwye
 2 years ago
Best ResponseYou've already chosen the best response.0Say 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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