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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?

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The answer is (a) 0.2s (b) 0.404m
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.
Since it was thrown with an angle why we need to consider the acceleration

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Other answers:

T=2usintheta/g can we use that??
hey is time taken =1sec
sorry i mean 2sec ??
and yea shakir u can use dat formula u mention above
How in question it is given that theta=30
lol am sorry I didn't read the question carefully. @heena will help you out :D
@heena i got the Time of Flight Plzz help me to do distance
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
so wats the time of flight am i right?? i mean my ans is correct??
but how?
i am also confused i did it with out taking any relative acceleration and velocity
still show me ur work wat u did
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??
and yea shakir for the second option u have to cal range means u^2sin2theta/g
not getting that
i the distance 0.2sqrt3 ??
do u knw sin2theta means 2xsinthetaxcostheta
but that is wrong
dude time will be the same in any case, if you consider inertial or a non inertial it will be 0.2 as T=2u sin theta/g
ok then wat abt range
but gautham dont u add lift's velocity too??
we dont have to ..think and tell
But not getting!!!!
Range should be 0.404 m somehow///
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?
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?
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
Any Idea Guys!
well i dont know how to do it in heena's way, although we can
wat abt Range!!!!!!!!!
im getting 0.044
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
so either time of flight isnt correct or shakir didnt type properly
no i am correct!
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
Time Wasting lets move to another question
ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a
this type of question is there in hc verma i remember doing it last 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
@ heena wont it be a-g?
  • 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?
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
@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
Try the second part :P
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?
@heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s
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
@heena - he has resolved the components
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 :)
@heena , You should not take the acceleration of lift into account. Think about it.
but y not @shivam_bhalla lift is not static u cant neglect that :O
@shivam_bhalla WAT ABT RANGE???
@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
@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
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
@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
HOW U=0.2?

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