kinematics
water drops fall at a constant rate from a nozzle fixed height h above the ground. when first drop hits the the ground, at that instant nth drop begins to fall. find distance of 2nd and 3rd drop from ground at this instant
idk the ans since i made this question myself.
33 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.you have to define the dimensions of the drop or else I don't think it will be possible to solve
idts thats necessary , i took this from kinematics mains pyq
oh yea ive seen alot of these
but i hate this
i never get it
chatgpt solved it well
the feelings mutual
dont trust ai
i checked it, everything made sense
I got an answer
$( \frac{2}{n}-\frac{3}{n^{2}} ) h$
Opt
i think the q asked individual distances of 2nd and 3rd drop from ground, not distance between then
yep
pls post your soln tho
Oops.
Yeah, let me do that
:sweaty:
gpt doesnt count the nth drop in sqrt(2h/g), i guess depends on the question's wording?
time between drops sqrt(2h/g) / n-1 nhi hoga? cuz nth drop release hui hai, not hit ground
It's time between releasing of drops
for example 4 drops hai..time between 1-2, 2-3, 3-4....3 intervals hue na
,rotate
this will be distance of 2nd drop from ground
u time of flight of kth drop as
time it took for first drop to reach ground - time it took for kth drop to move from nozzle
right?
Yes
:sweaty: I am getting confused now
yeah intervals n-1 hue but we are concerned about total time , total drops and rate of falling
n-1 intervals ko hi toh total time i.e, root 2 h/g laga
hm i mean we hv to find time between releasing drops right...say n=4, so first drop and second drop me 't' time ka diff tha..then 2nd and 3rd me bhi 't' time...3rd and 4th drop me bhi 't' time tha...so 4th drop jab release hui tab 3 time intervals the na...and we know that tab first drop reached ground..so 3t = root 2h/g
hm makes sense
👍
so the time travelled for 2nd drop is root 2h/g - root 2h/g(n-1)?
t = sqrt(2h/g) (time for one drop to reach the ground)
r = n-1/t (rate of drops coming out of the nozzle)
time at which the 2nd drop falls = 1/r
time at which the 3rd drop falls = 2/r
distance travelled from top = g/2 (t)^2
=> from ground = h - g/2 (t)^2
time of flight of 2nd drop = t-t/n-1
time of flight of 3rd drop = t- 2t/n-1
yes ig
prolly equate that and subtract from h