2D moshun
no idea how to even approach this ... i feel that the question giving us "hits the plane perpendicularly" has some use just that WHAT IS TRYING TO TELL US
52 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.just think about it , it hits the plane perpendiculalrly
and also the plane is at angle 45 degree , if u resolve the speed component at that uh plane part Vx = Vy try thinking from there
Vx = Vy by resolving the vector Vcos45 and Vsin45 and they are equal
at the instant it is about to hit the plane???
nahh didnt get your point
ok one approach i did get was applying Kinematics equation for Vy
but i would need T
and T needs Uy
but Uy is 0
š
cant use Ux wo to dhundhna hai
Ux=Vx=Vy
let it fall height h then Vy= root(2gh)
oh yea
and length of plank = (10-h)cos45
for about motion under gravity
ok ji
ill try once
aaye toh batana i am not sure too
bhai time of flight be giving problem
ok as a move in desperation i took a look at the solution
and they resolved the axis along the plane(why did i not do that)
and then did all the calculation taking g ka components and all
gonna proceed with that
bruh this makes no sense
idk im gonna go to eat smh i cant think straight
uss se aa gya?
That's legit what I said
Wait nvm
Yes
idk
i was down
to eat
then got to work for tommrow(school jaana hai š )
ab leke baitha hu wapis
give 10-15 mins
its been a while since i picked up 2D
wel it wasnt understandable bhai tune likha kya tha matlab kya tha š
did that idk what i was doing but nahi ha
yupp 18 ho gaya
i meant to say that when you resolve vector at the point when particle is touching plane , Vx= Vy so thats an extra equation which helps ig
yea yea absolutely correct dude
ok so i got u
so if i want
OB
i can say that OB will be the displacement along X in time t
t= u/g
but i am not getting the anser
no
thats AB , displacement along x in time T , and its not even AB tbh it goes down as well
wat wat
wait
uhh
its
uh
wait
10cos45 + the range of the projective from initial to the final point along the X direction of incline
just a sec
X is along the incline
getting 25/root(2)
what am i doing wrong
you can try one thing idk if it will work , when u break the component at the plane and vx = vy
find out how much time it took
u/g
then u can use that to calc the distance subtract that from 10 and u get hieght
and then use trigo to get OB
one sec
so
basically
calc the distance it travlled down , subtract that from 10
and use trigo
okk
gib minutes
wait wdym travelled down???
travelled down
along the incline like
uhhh
man i have to clikc photos then send to laptop and then upload here
@Gamertug
this>
??
yes calc that
and subtract from 10
bro how
to get height
everything i have
is along invline
thi isnt along incline
why would i create new values for the same sub question
what
like
we know "u" and at the point it touches plane Vy = u
Vi = 0
a = g
we can find s
ayo why u=0???
u=v no??
it is already being shot with some velocity
how can we take 0
sorry i meant (Vi)y = 0
oh ok
we applying eqn along y
yea yea one sec
bhai nahi aaya š lemme send my working 2 mins
oh wait
wrong value liya maine
nahi aaya to phir bhi nahi
lemme try and solve
thiss
u = 10root 3/2 is correct ?
now that i see it uhm
its 10root 2/3
im an idiot
....
isliye mere number nahi aate
same bhai bahut silly mistake hota hai
;-;
aa gaya
thanku ji
i sometimes wonder
nahh chodo
thanks @Gamertug @
what
np
@hardcoreisdead
+solved @Gamertug @hardcoreisdead
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