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?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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?

IIT study group
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

this is already in the physics group
Can u Check My answer i got h = 148.48 m

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
thanks!!
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....
thxxxx
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! :|
This one looks correct
But This one Luks Similar to 1st One @shubhamsrg
1st one ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question