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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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shakirBest ResponseYou've already chosen the best response.0
The answer is (a) 0.2s (b) 0.404m
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
relative initial v = 0.2 m/sec relative acc. = g0.2 m/sec^2 cal. final vel. now using newton's equations of motions.
 one year ago

shakirBest ResponseYou've already chosen the best response.0
Since it was thrown with an angle why we need to consider the acceleration
 one year ago

shakirBest ResponseYou've already chosen the best response.0
T=2usintheta/g can we use that??
 one year ago

heenaBest ResponseYou've already chosen the best response.0
and yea shakir u can use dat formula u mention above
 one year ago

shakirBest ResponseYou've already chosen the best response.0
How in question it is given that theta=30
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
lol am sorry I didn't read the question carefully. @heena will help you out :D
 one year ago

shakirBest ResponseYou've already chosen the best response.0
@heena i got the Time of Flight Plzz help me to do distance
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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
 one year ago

heenaBest ResponseYou've already chosen the best response.0
so wats the time of flight am i right?? i mean my ans is correct??
 one year ago

shakirBest ResponseYou've already chosen the best response.0
i am also confused i did it with out taking any relative acceleration and velocity
 one year ago

heenaBest ResponseYou've already chosen the best response.0
still show me ur work wat u did
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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??
 one year ago

heenaBest ResponseYou've already chosen the best response.0
and yea shakir for the second option u have to cal range means u^2sin2theta/g
 one year ago

heenaBest ResponseYou've already chosen the best response.0
i the distance 0.2sqrt3 ??
 one year ago

heenaBest ResponseYou've already chosen the best response.0
do u knw sin2theta means 2xsinthetaxcostheta
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
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
 one year ago

heenaBest ResponseYou've already chosen the best response.0
but gautham dont u add lift's velocity too??
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
we dont have to ..think and tell
 one year ago

shakirBest ResponseYou've already chosen the best response.0
Range should be 0.404 m somehow///
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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?
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
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?
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
well i dont know how to do it in heena's way, although we can
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
so either time of flight isnt correct or shakir didnt type properly
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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
 one year ago

shakirBest ResponseYou've already chosen the best response.0
Time Wasting lets move to another question
 one year ago

heenaBest ResponseYou've already chosen the best response.0
ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
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
 one year ago

phiBest ResponseYou've already chosen the best response.0
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
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
@ heena wont it be ag?
 one year ago

phiBest ResponseYou've already chosen the best response.0
also, there is ambiguity when stating the velocity of the stone: is it relative to the elevator floor (apparently) or relative to the earth?
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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 ga yea u are right :P
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@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
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
Try the second part :P
 one year ago

shakirBest ResponseYou've already chosen the best response.0
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
 one year ago

heenaBest ResponseYou've already chosen the best response.0
means time of flight is =2usintheta/g = 2x 0.4 x 1/2x 9.6 =o.o41 sec i m gettin @gautham and @bhalla u are given parabolic path way so how can u apply vertical motion formula?
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s
 one year ago

heenaBest ResponseYou've already chosen the best response.0
no i didnt retriceume g=10 as lift i moving upward means ga that is 9.80.2=9.6 and u = 0.4m/sec as lift velocity is also added
 one year ago

goutham1995Best ResponseYou've already chosen the best response.0
@heena  he has resolved the components
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@heena , You should not take the acceleration of lift into account. Think about it.
 one year ago

heenaBest ResponseYou've already chosen the best response.0
but y not @shivam_bhalla lift is not static u cant neglect that :O
 one year ago

shakirBest ResponseYou've already chosen the best response.0
@shivam_bhalla WAT ABT RANGE???
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@shakir , why do you require the range ?? @heena , assume you are standing in a train compartment and throwing a projectile in a compartment. Does the acceleration of the ball change in any way for the projectile with respect to the train. The answer is no. The same applies for elevator as well
 one year ago

shakirBest ResponseYou've already chosen the best response.0
PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@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
 one year ago

shakirBest ResponseYou've already chosen the best response.0
YES THEN THE DISTANCE IS ???????
 one year ago

heenaBest ResponseYou've already chosen the best response.0
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
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
@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
 one year ago
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