anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The answer is (a) 0.2s (b) 0.404m
apoorvk
  • 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
  • anonymous
Since it was thrown with an angle why we need to consider the acceleration

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
T=2usintheta/g can we use that??
anonymous
  • anonymous
hey is time taken =1sec
anonymous
  • anonymous
sorry i mean 2sec ??
anonymous
  • anonymous
and yea shakir u can use dat formula u mention above
anonymous
  • anonymous
How in question it is given that theta=30
apoorvk
  • apoorvk
lol am sorry I didn't read the question carefully. @heena will help you out :D
anonymous
  • anonymous
@heena i got the Time of Flight Plzz help me to do distance
anonymous
  • 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
  • anonymous
so wats the time of flight am i right?? i mean my ans is correct??
anonymous
  • anonymous
0.2s
anonymous
  • anonymous
but how?
anonymous
  • anonymous
i am also confused i did it with out taking any relative acceleration and velocity
anonymous
  • anonymous
still show me ur work wat u did
anonymous
  • anonymous
|dw:1338466097736:dw|
anonymous
  • 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
  • anonymous
cany=*can
anonymous
  • anonymous
and yea shakir for the second option u have to cal range means u^2sin2theta/g
anonymous
  • anonymous
not getting that
anonymous
  • anonymous
i the distance 0.2sqrt3 ??
anonymous
  • anonymous
do u knw sin2theta means 2xsinthetaxcostheta
anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1338466702370:dw|
anonymous
  • anonymous
but that is wrong
anonymous
  • 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
  • anonymous
ok then wat abt range
anonymous
  • anonymous
but gautham dont u add lift's velocity too??
anonymous
  • anonymous
we dont have to ..think and tell
anonymous
  • anonymous
But not getting!!!!
anonymous
  • anonymous
Range should be 0.404 m somehow///
anonymous
  • 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
  • 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
  • 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
  • anonymous
Any Idea Guys!
anonymous
  • anonymous
well i dont know how to do it in heena's way, although we can
anonymous
  • anonymous
wat abt Range!!!!!!!!!
anonymous
  • anonymous
im getting 0.044
anonymous
  • 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
  • anonymous
so either time of flight isnt correct or shakir didnt type properly
anonymous
  • anonymous
no i am correct!
anonymous
  • 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
  • anonymous
Confused!!
anonymous
  • anonymous
Time Wasting lets move to another question
anonymous
  • anonymous
ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a
anonymous
  • 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
  • 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
  • anonymous
@ heena wont it be a-g?
anonymous
  • anonymous
@heena *
phi
  • 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
  • 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
  • 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
  • anonymous
Try the second part :P
anonymous
  • 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
@shivam_bhalla
anonymous
  • anonymous
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?
anonymous
  • anonymous
@heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s
anonymous
  • 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
  • anonymous
@heena - he has resolved the components
anonymous
  • 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
  • anonymous
@heena , You should not take the acceleration of lift into account. Think about it.
anonymous
  • anonymous
but y not @shivam_bhalla lift is not static u cant neglect that :O
anonymous
  • anonymous
@shivam_bhalla WAT ABT RANGE???
anonymous
  • anonymous
@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
  • anonymous
PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla
anonymous
  • 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
  • anonymous
YES THEN THE DISTANCE IS ???????
anonymous
  • anonymous
@shivam_bhalla
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
  • 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
  • anonymous
HOW U=0.2?

Looking for something else?

Not the answer you are looking for? Search for more explanations.