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anonymous
 4 years 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?
anonymous
 4 years 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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer is (a) 0.2s (b) 0.404m

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since it was thrown with an angle why we need to consider the acceleration

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0T=2usintheta/g can we use that??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey is time taken =1sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and yea shakir u can use dat formula u mention above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How in question it is given that theta=30

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@heena i got the Time of Flight Plzz help me to do distance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0so wats the time of flight am i right?? i mean my ans is correct??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am also confused i did it with out taking any relative acceleration and velocity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0still show me ur work wat u did

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338466097736:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 4 years ago
Best ResponseYou've already chosen the best response.0and yea shakir for the second option u have to cal range means u^2sin2theta/g

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i the distance 0.2sqrt3 ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do u knw sin2theta means 2xsinthetaxcostheta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338466702370:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dude 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok then wat abt range

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but gautham dont u add lift's velocity too??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we dont have to ..think and tell

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Range should be 0.404 m somehow///

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah..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
 4 years ago
Best ResponseYou've already chosen the best response.0but 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
 4 years ago
Best ResponseYou've already chosen the best response.0well i dont know how to do it in heena's way, although we can

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wat abt Range!!!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u 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
 4 years ago
Best ResponseYou've already chosen the best response.0so either time of flight isnt correct or shakir didnt type properly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea 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
 4 years ago
Best ResponseYou've already chosen the best response.0Time Wasting lets move to another question

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0@ heena wont it be ag?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try the second part :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0means 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@heena  he has resolved the components

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah 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
 4 years ago
Best ResponseYou've already chosen the best response.0@heena , You should not take the acceleration of lift into account. Think about it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but y not @shivam_bhalla lift is not static u cant neglect that :O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@shivam_bhalla WAT ABT RANGE???

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YES THEN THE DISTANCE IS ???????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no @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
 4 years ago
Best 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
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