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Yahoo!
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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?
 one year ago
 one year ago
Yahoo! Group Title
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?
 one year ago
 one year ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
@experimentX @shubhamsrg @JFraser
 one year ago

JFraser Group TitleBest ResponseYou've already chosen the best response.0
this is already in the physics group
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
Can u Check My answer i got h = 148.48 m
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
s = 1/2 gt^2  v = 0 + g(t1) now u = v then, 1/3 s = v(1) + 1/2 g(1)^2 s = 1/2 gt^2  from these two solve or 's' and 't' rest should be okay
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
nicely explained @experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
thanks!!
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
Ok....i Did by this way: h = 1/2 g t^2 Dn = u + a/2 (2n1) h/3 = a/2(2t1).....?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
in the formula you used @Yahoo! 'n' is the nth second, in your case, n= t1 .. you already know h= 1/2 gt^2 also h/3 =g/2 (2t 3) taking g=10, i get h=45..someone please confirm ?
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
But why Do we take t1
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
last second right ? after t seconds are complete , the object doesnt travel further..
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
I am Silly....u r correct...i got it now....
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
i may be wrong in logic though.. let someone else make any correction if its there..
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
I think ur Logic is Correct...
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
But Lets Confirm..as u told..@sauravshakya
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
maybe not tagged correctly..allow me @sauravshakya
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
you got 148 right? by another plausible method, i am also getting 148, so that should be right,,my logic is thus, flawed//hmm..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
here is what i did,, we h = 1/2 g t1 ^2 ....(1) where t1 is total time taken . let t2 be time before it comes to last second of the journey, thus t1  t2 =1 => t2 = t1 1....(2) we know h h/3 = 1/2 g t2 ^2 2h/3 = 1/2 g t2 ^2 ...(3) using 1,2,and 3 eqns, we can solve for h, t1 and t2 .. by taking g=10, i approximately get h=148.4 something.. which should be the ans.. unless ofcorse,,there is some flaw in this also! :
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
This one looks correct
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
But This one Luks Similar to 1st One @shubhamsrg
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
1st one ?
 one year ago
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