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shakir Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. shakir Group Title
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    The answer is (a) 0.2s (b) 0.404m

    • 2 years ago
  2. apoorvk Group Title
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    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.

    • 2 years ago
  3. shakir Group Title
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    Since it was thrown with an angle why we need to consider the acceleration

    • 2 years ago
  4. shakir Group Title
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    T=2usintheta/g can we use that??

    • 2 years ago
  5. heena Group Title
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    hey is time taken =1sec

    • 2 years ago
  6. heena Group Title
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    sorry i mean 2sec ??

    • 2 years ago
  7. heena Group Title
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    and yea shakir u can use dat formula u mention above

    • 2 years ago
  8. shakir Group Title
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    How in question it is given that theta=30

    • 2 years ago
  9. apoorvk Group Title
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    lol am sorry I didn't read the question carefully. @heena will help you out :D

    • 2 years ago
  10. shakir Group Title
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    @heena i got the Time of Flight Plzz help me to do distance

    • 2 years ago
  11. heena Group Title
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    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

    • 2 years ago
  12. heena Group Title
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    so wats the time of flight am i right?? i mean my ans is correct??

    • 2 years ago
  13. shakir Group Title
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    0.2s

    • 2 years ago
  14. heena Group Title
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    but how?

    • 2 years ago
  15. shakir Group Title
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    i am also confused i did it with out taking any relative acceleration and velocity

    • 2 years ago
  16. heena Group Title
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    still show me ur work wat u did

    • 2 years ago
  17. shakir Group Title
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    |dw:1338466097736:dw|

    • 2 years ago
  18. heena Group Title
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    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??

    • 2 years ago
  19. heena Group Title
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    cany=*can

    • 2 years ago
  20. heena Group Title
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    and yea shakir for the second option u have to cal range means u^2sin2theta/g

    • 2 years ago
  21. shakir Group Title
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    not getting that

    • 2 years ago
  22. heena Group Title
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    i the distance 0.2sqrt3 ??

    • 2 years ago
  23. heena Group Title
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    do u knw sin2theta means 2xsinthetaxcostheta

    • 2 years ago
  24. shakir Group Title
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    yes

    • 2 years ago
  25. shakir Group Title
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    |dw:1338466702370:dw|

    • 2 years ago
  26. shakir Group Title
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    but that is wrong

    • 2 years ago
  27. goutham1995 Group Title
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    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

    • 2 years ago
  28. shakir Group Title
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    ok then wat abt range

    • 2 years ago
  29. heena Group Title
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    but gautham dont u add lift's velocity too??

    • 2 years ago
  30. goutham1995 Group Title
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    we dont have to ..think and tell

    • 2 years ago
  31. shakir Group Title
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    But not getting!!!!

    • 2 years ago
  32. shakir Group Title
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    Range should be 0.404 m somehow///

    • 2 years ago
  33. heena Group Title
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    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?

    • 2 years ago
  34. goutham1995 Group Title
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    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?

    • 2 years ago
  35. heena Group Title
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    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

    • 2 years ago
  36. shakir Group Title
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    Any Idea Guys!

    • 2 years ago
  37. goutham1995 Group Title
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    well i dont know how to do it in heena's way, although we can

    • 2 years ago
  38. shakir Group Title
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    wat abt Range!!!!!!!!!

    • 2 years ago
  39. goutham1995 Group Title
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    im getting 0.044

    • 2 years ago
  40. heena Group Title
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    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

    • 2 years ago
  41. goutham1995 Group Title
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    so either time of flight isnt correct or shakir didnt type properly

    • 2 years ago
  42. shakir Group Title
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    no i am correct!

    • 2 years ago
  43. heena Group Title
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    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

    • 2 years ago
  44. shakir Group Title
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    Confused!!

    • 2 years ago
  45. shakir Group Title
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    Time Wasting lets move to another question

    • 2 years ago
  46. heena Group Title
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    ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a

    • 2 years ago
  47. goutham1995 Group Title
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    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

    • 2 years ago
  48. phi Group Title
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    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

    • 2 years ago
  49. goutham1995 Group Title
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    @ heena wont it be a-g?

    • 2 years ago
  50. goutham1995 Group Title
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    @heena *

    • 2 years ago
  51. phi Group Title
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    also, there is ambiguity when stating the velocity of the stone: is it relative to the elevator floor (apparently) or relative to the earth?

    • 2 years ago
  52. heena Group Title
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    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

    • 2 years ago
  53. shivam_bhalla Group Title
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    @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

    • 2 years ago
  54. shivam_bhalla Group Title
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    Try the second part :P

    • 2 years ago
  55. shakir Group Title
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    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

    • 2 years ago
  56. shakir Group Title
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    @shivam_bhalla

    • 2 years ago
  57. heena Group Title
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    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?

    • 2 years ago
  58. shivam_bhalla Group Title
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    @heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s

    • 2 years ago
  59. heena Group Title
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    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

    • 2 years ago
  60. goutham1995 Group Title
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    @heena - he has resolved the components

    • 2 years ago
  61. heena Group Title
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    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 :)

    • 2 years ago
  62. shivam_bhalla Group Title
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    @heena , You should not take the acceleration of lift into account. Think about it.

    • 2 years ago
  63. heena Group Title
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    but y not @shivam_bhalla lift is not static u cant neglect that :O

    • 2 years ago
  64. shakir Group Title
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    @shivam_bhalla WAT ABT RANGE???

    • 2 years ago
  65. shivam_bhalla Group Title
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    @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

    • 2 years ago
  66. shakir Group Title
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    PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

    • 2 years ago
  67. shivam_bhalla Group Title
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    @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

    • 2 years ago
  68. shakir Group Title
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    YES THEN THE DISTANCE IS ???????

    • 2 years ago
  69. shakir Group Title
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    @shivam_bhalla

    • 2 years ago
  70. heena Group Title
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    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

    • 2 years ago
  71. shivam_bhalla Group Title
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    @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

    • 2 years ago
  72. shakir Group Title
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    HOW U=0.2?

    • 2 years ago
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