An elevator is going up with an upward acceleration of 0.2m/s^2 . At the istant when the velocity is 0.2m/s a stone is projected upwards from its floor with a speed of 2m/s at an angle 30 degree with horizontal, relative to the elevator.
Find (a) The time taken by the stone to return to the floor
(b) The distance travelled by the floor of the lift from the time the stone was projected to the time when the stone hits the floor is?

- anonymous

- jamiebookeater

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- anonymous

The answer is (a) 0.2s (b) 0.404m

- apoorvk

relative initial v = 0.2 m/sec
relative acc. = -g-0.2 m/sec^2
cal. final vel. now using newton's equations of motions.

- anonymous

Since it was thrown with an angle why we need to consider the acceleration

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## More answers

- anonymous

T=2usintheta/g can we use that??

- anonymous

hey is time taken =1sec

- anonymous

sorry i mean 2sec ??

- anonymous

and yea shakir u can use dat formula u mention above

- anonymous

How in question it is given that theta=30

- apoorvk

lol am sorry I didn't read the question carefully. @heena will help you out :D

- anonymous

@heena i got the Time of Flight Plzz help me to do distance

- anonymous

so u knw already lift imoving means tone already have velocity 2m/s with it now u also thrown it by 2m/s velocity so total velocity becomes 4m/s
now we know qn is like this
|dw:1338465698465:dw|
here u knw acceleration of lift is 0.2m/s^2 means it will be same for stone too now u have u=4m/sec u have theta 3odegree and here g will be taken as 0.2m/s^2 nw put the values in formula u ll get it

- anonymous

so wats the time of flight am i right?? i mean my ans is correct??

- anonymous

0.2s

- anonymous

but how?

- anonymous

i am also confused i did it with out taking any relative acceleration and velocity

- anonymous

still show me ur work wat u did

- anonymous

|dw:1338466097736:dw|

- anonymous

hmm but here velocity shouldnt be 4 ?? cany anyone xplain me as i read when an obeject is in the moving body the velocity of moving body is also added to the object??

- anonymous

cany=*can

- anonymous

and yea shakir for the second option u have to cal range means u^2sin2theta/g

- anonymous

not getting that

- anonymous

i the distance 0.2sqrt3 ??

- anonymous

do u knw sin2theta means 2xsinthetaxcostheta

- anonymous

yes

- anonymous

|dw:1338466702370:dw|

- anonymous

but that is wrong

- anonymous

dude time will be the same in any case, if you consider inertial or a non inertial frame..so it will be 0.2 as T=2u sin theta/g

- anonymous

ok then wat abt range

- anonymous

but gautham dont u add lift's velocity too??

- anonymous

we dont have to ..think and tell

- anonymous

But not getting!!!!

- anonymous

Range should be 0.404 m somehow///

- anonymous

but a far i remember for example if we are in moving train a boy in the trains throws a ball in a forwards direction that time we also add train's velocity along with the velocity of by which ball is thrown isntd?

- anonymous

yeah..but time at which the ball falls for an observer outside the train is the same as the time when the ball falls for an observer inside the train right?

- anonymous

but here acc to qn we have to add it too right but answer i coming without adding it i m not getting any valuable reason for not adding their velocity

- anonymous

Any Idea Guys!

- anonymous

well i dont know how to do it in heena's way, although we can

- anonymous

wat abt Range!!!!!!!!!

- anonymous

im getting 0.044

- anonymous

u knw shakir i m gettin range = 0.364 :( so i also dunno as i m not gettin how to work here i mean u dunno the total ditance the lift have to travel right and if u see at 0.2 ec the lift travel s=ut + 1/2at^2
u ll get 0.044m

- anonymous

so either time of flight isnt correct or shakir didnt type properly

- anonymous

no i am correct!

- anonymous

yea dats y i m asking from u guys he is only trieing to get the ans no matter how u are getting u knw he is using hit and trial method.. but i wanna knw the real method

- anonymous

Confused!!

- anonymous

Time Wasting lets move to another question

- anonymous

ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a

- anonymous

this type of question is there in hc verma i remember doing it last year..so you can go thru that to get the method

- phi

I think there is a typo in your answer key. The vertical distance traveled by the elevator is 0.044 m
not 0.404 m

- anonymous

@ heena wont it be a-g?

- anonymous

@heena *

- phi

also, there is ambiguity when stating the velocity of the stone: is it relative to the elevator floor (apparently) or relative to the earth?

- anonymous

no as i ermeber when i use lift and we move down i feel heavy means g+a and when we move upward i feel light means den it will be g-a yea u are right :P

- anonymous

@shakir , It must be easy enough.
For y axis
\[v_y = u_y - (g)t\]
\[u_y = 2 \sin(30^{0})=1\]
\[v_y = 1 -(g)t\]
At maximum height, v_y = 0
0 = 1- 10 t
t =1/10
Time taken to reach maximum height = 1/10 sec
Therefore time of flight = 2 * 1/10= 0.2 sec

- anonymous

Try the second part :P

- anonymous

I TRIED THE SECON PART BUT I GOT IT AS 0.364 M BUT THE ANSWER IS 0.404M U CAN SEE MY WORK ABOVE

- anonymous

- anonymous

@heena , mistake 1 = assume g = 10 m/s
mistake 2 = u = 2 m/s

- anonymous

no i didnt retriceume g=10 as lift i moving upward means g-a that is 9.8-0.2=9.6
and u = 0.4m/sec as lift velocity is also added

- anonymous

@heena - he has resolved the components

- anonymous

ah leave it i ll ask this to my teacher :P he will xplain me v.well and clear my doubt by telling my mistake anyways thankyou guys for helping me :)

- anonymous

@heena , You should not take the acceleration of lift into account. Think about it.

- anonymous

but y not @shivam_bhalla lift is not static u cant neglect that :O

- anonymous

@shivam_bhalla WAT ABT RANGE???

- anonymous

- anonymous

PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

- anonymous

@shakir , read the second part of the question carefully. It doesn't aks the range of projectile. It asks about the distance travelled by the floor of the lift during the time of flight

- anonymous

YES THEN THE DISTANCE IS ???????

- anonymous

- anonymous

no @shivam_bhalla i did such qn and if u practically think abu this den my statement is correct dat we cant neglect lift's acclereation and as i said dont xplain me anymore i m gettin more confuse and by ur statement i m having more doubts so better i ll eat my teacher's brain :P :P so i ll ans tmrw :P

- anonymous

@shakir
Apply.
v= u + at
t = 0.2
u = 0.2
Think about what your acceleration must be.
GTG now. Will check this thread later

- anonymous

HOW U=0.2?

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