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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s = 1/2 gt^2
v = 0 + g(t-1) now u = v
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
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)
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! :|