Question

A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is 12.8 m above the water surface, the fish sees the speed of ball as

Answer 1

@Mashy dude.. :D

Answer 2

@ParthKohli

Answer 3

Ek baat bataun?

Answer 4

Mujhe Physics nahi aati... tag mat karo >_>

Answer 5

Tereko sab aata hai. Mereko pta hai. :)

Answer 6

Nahi aata na!

Answer 7

Ok. :/

Answer 8

hmmm lemme think!

Answer 9

Ho gya. :D

Answer 10

yea .. i got it too! :D

Answer 11

How did you do it btw. Just to see if your method is easier.

Answer 12

no no first check whether my method is correct :P.. lemme type

Answer 13

First find the height of the ball as seen by the fish when it is 20m above and when it is 12.8m above .. using normal shift formula

then find the difference .. that gives the apparent displacement of the ball as seen by the fish

then use vsquare = 2gh
where h is the calculated apparent displacement?

Answer 14

Wait. I did that too. I think i solved it wrong.

Answer 15

m high on pizza.. eating my 7th piece.. so cut me some slack if i do some mistake :D

Answer 16

x-ap=80/3?

Answer 17

I mean x(ap). Not x (minus) ap . :P

Answer 18

And no i didnt get it. v=sqrt2gh=12m/s
Then using the shift formula. x(image of the ball)=4/3 x(ball) ..
What next?

Answer 19

Hmm. So differntiating both sides wrt time gives me the answer. I am not sure if that is the correct approach.

Answer 20

wait..

aph = 20* 3/4 = 15m
and aph2 = 12.8*3/4= 9.6

hence displacement = 15 - 9.6 = 5.4

hence v square = 2*9.8*5.4

that doesn't give the answer?

Answer 21

Nope.

Answer 22

ahh.. so m sure it sees that the ball has travelled that much distance .. hmmmm

Answer 23

What? :o

Answer 24

arre.. the fish sees the ball to be having that much displacement only na??? 5.4m displacement?! .. but i think it the acceleration no longer remains g.. yea!!.. things get screwed up..

Answer 25

Why doesnt acc remain g?

Answer 26

cause everything is normal shifted.. and the normal shift depends on the depth of the object as well.. so as the object keeps coming closer and closer, the Normal shift keeps decreasing... hence no longer is it accelerating at g .. i dun even know if the acceleration remains constant :P

Answer 27

Ok so differentiating h(ap)=h(actual)*myu, seems logical?

Answer 28

wrt time...

Answer 29

hmm.. yea..so all you do is find the actual velocity and multiply by myu? and besides i did it wrong earlier anyways :D

Answer 30

Haha.. Another question coming up. i'll upload. Wait.

Answer 31

wait.. so is taht it??

Answer 32

Yup. That gives me the answer.

Answer 33

its definitely not 1 and 4 for sure

Answer 34

it can't be 3.. also!!

Answer 35

i t has to be 2

Answer 36

Yes, the critical angle thing. Why not 3?

Answer 37

because in three.. the transmission is at 100%.. if that was the case.. the reflected ray should have been at zero percent but its not.. so its impossible.. !!

Answer 38

initially i mean!!

Answer 39

:O
I get it...
Thank you :D

Answer 40

ok .. so the medal is for THIS question.. and not the previous one.. :D

Answer 41

Haha yeah. :D

Answer 42

More questions tomorrow. :P

Answer 43

are you into dbz??

Answer 44

that is for the earlier qusetion :D

Answer 45

dbz?
And thank you.

Answer 46

dragon ball z?

Answer 47

Oh, i used to love it. But not now. I dont like watching television anyway.