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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Stacey Warren - Expert brainly.com
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s = 1/2 gt^2 ----------- v = 0 + g(t-1) 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
nicely explained @experimentX
Ok....i Did by this way: h = 1/2 g t^2 Dn = u + a/2 (2n-1) h/3 = a/2(2t-1).....?
in the formula you used @Yahoo! 'n' is the nth second, in your case, n= t-1 .. 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 ?
But why Do we take t-1
last second right ? after t seconds are complete , the object doesnt travel further..
I am Silly....u r correct...i got it now....
i may be wrong in logic though.. let someone else make any correction if its there..
I think ur Logic is Correct...
But Lets Confirm..as u told..@sauravshakya
maybe not tagged correctly..allow me @sauravshakya
you got 148 right? by another plausible method, i am also getting 148, so that should be right,,my logic is thus, flawed//hmm..
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! :|