Question

formula derivation!! i kinda confused T_T

Answer 1

why they are not considering x???

Answer 2

@matt101 @ParthKohli @mathslover @experimentX

Answer 3

for derivation i used this formula
\[s = ut+ \frac{ 1 }{ 2 }at ^{2}\]

Answer 4

v^2 = u^2 + 2 a s
v = u + at
use this formulas.

Answer 5

s = ut + 1/2 at^2

Answer 6

but i don't have v

Answer 7

yeah i used second one

Answer 8

how did you calculate the v??

Answer 9

i did not

Answer 10

i don't need v

Answer 11

or i do??

Answer 12

this is what i did

Answer 13

Apply S=ut+1/2at^2 from 0 to M1
then u get one equation-----
x= 0.5at1^2------1

then apply S=ut + 1/2 at^2 from 0 to M2
then u get another equation---

x+h = 0.5at2^2-------2
substitute x from eq 1 in eq 2
u get the answer

Answer 14

i think there has been a slight mistake in ur equation
where displacement is x+h then the time should be (t1+t2)^2
rest seems ok

Answer 15

ok im confused as hell

Answer 16

@No.name

Answer 17

hey acceleration is constant,why cant u solve it now

Answer 18

@thushananth01 so i did that thingy actually wait see my work

Answer 19

i dunno if u'll get it but check it out

Answer 20

what do u mean @aajugdar ??

Answer 21

You need to apply all three kinematic equations of motion over here ,

Answer 22

oh wait
his equation looks correct,i think t2 isnt time from m1 but x

Answer 23

There is uniform acceleration , so no worries,
If it was not uniform u had to use integration (calculus)

Answer 24

u cant apply between t1 and t2 cuz they initial velocity at t1, so its better if u apply from 0 to M1 and 0 to M2
@aajugdar T2 is the time taken for the ball to travel from 0 to M2 so i dont find anythng wrong in my equation :/

Answer 25

yes it is correct
my bad

Answer 26

Shall I help?

Answer 27

okay wait so i use that equation to find a from 0 to t1 which is distance of x
and from 0 to t2 which is distance h

Answer 28

IS the answer D

Answer 29

distance is x + h @Somy

Answer 30

yes

Answer 31

its D

Answer 32

@Somy did u get it

Answer 33

first of all ...the height travelled in first t1 time is..

x = o + 1/2 a*(t1)^2....(as the body starts from rest)...............{Equation 1}

and from the rest position in time t2 the ball fall a height of (h+x)..

so

h+x = 0+ 1/2 a*(t2)^2....................................eqn (II)
and we know the change in height from t1 to t2 is h ..so h should be given by subtracting the right hand expression of both the eqns...
so
h = [1/2 a*(t2)^2 ] -[1/2 a*(t1)^2]

h= 1/2 a* [ (t2)^2 - (t1)^2]

...now solve the eqn and u will get a = 2h/[ (t2)^2 -(t1)^2]....so answer is D.

Answer 34

So the answer is D

Answer 35

wait if distance is h + x then time would be t2-t1

Answer 36

okay let me see it @No.name

Answer 37

YUP!

Answer 38

oh!! wait i do simultaneous equation???

Answer 39

@No.name @thushananth01

Answer 40

Yes try it and see!!

Answer 41

why t2- t1? they have given tht the time taken from 0 to m2 is t2 (thats why its t2)

Answer 42

okay gimme some time i'll try

Answer 43

yaaaaaaaaaaay thank u guys sooo much!!!!
@No.name @thushananth01 @aajugdar @experimentX

Answer 44

No problem!

Answer 45

happy to be of help!

Answer 46

:D

Answer 47

i gave u a medal in some other q u asked since i appreciate your help!!! @No.name

Answer 48

thank you !

Answer 49

no need :D u deserved it :)

Answer 50

Much obliged @No.Name