Question

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?

Answer 1

@experimentX @shubhamsrg @JFraser

Answer 2

this is already in the physics group

Answer 3

Can u Check My answer i got h = 148.48 m

Answer 4

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

Answer 5

nicely explained @experimentX

Answer 6

thanks!!

Answer 7

Ok....i Did by this way:

h = 1/2 g t^2

Dn = u + a/2 (2n-1)

h/3 = a/2(2t-1).....?

Answer 8

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 ?

Answer 9

But why Do we take t-1

Answer 10

last second right ?
after t seconds are complete , the object doesnt travel further..

Answer 11

I am Silly....u r correct...i got it now....

Answer 12

thxxxx

Answer 13

i may be wrong in logic though.. let someone else make any correction if its there..

Answer 14

I think ur Logic is Correct...

Answer 15

But Lets Confirm..as u told..@sauravshakya

Answer 16

maybe not tagged correctly..allow me
@sauravshakya

Answer 17

you got 148 right?
by another plausible method, i am also getting 148, so that should be right,,my logic is thus, flawed//hmm..

Answer 18

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! :|

Answer 19

This one looks correct

Answer 20

But This one Luks Similar to 1st One @shubhamsrg

Answer 21

1st one ?