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shakir

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

    • one year ago
  2. apoorvk
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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.

    • one year ago
  3. shakir
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    Since it was thrown with an angle why we need to consider the acceleration

    • one year ago
  4. shakir
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    T=2usintheta/g can we use that??

    • one year ago
  5. heena
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    hey is time taken =1sec

    • one year ago
  6. heena
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    sorry i mean 2sec ??

    • one year ago
  7. heena
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    and yea shakir u can use dat formula u mention above

    • one year ago
  8. shakir
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    How in question it is given that theta=30

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

    • one year ago
  10. shakir
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    @heena i got the Time of Flight Plzz help me to do distance

    • one year ago
  11. heena
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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

    • one year ago
  12. heena
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    so wats the time of flight am i right?? i mean my ans is correct??

    • one year ago
  13. shakir
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    0.2s

    • one year ago
  14. heena
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    but how?

    • one year ago
  15. shakir
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    i am also confused i did it with out taking any relative acceleration and velocity

    • one year ago
  16. heena
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    still show me ur work wat u did

    • one year ago
  17. shakir
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    |dw:1338466097736:dw|

    • one year ago
  18. heena
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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??

    • one year ago
  19. heena
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    cany=*can

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

    • one year ago
  21. shakir
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    not getting that

    • one year ago
  22. heena
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    i the distance 0.2sqrt3 ??

    • one year ago
  23. heena
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    do u knw sin2theta means 2xsinthetaxcostheta

    • one year ago
  24. shakir
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    yes

    • one year ago
  25. shakir
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    |dw:1338466702370:dw|

    • one year ago
  26. shakir
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    but that is wrong

    • one year ago
  27. goutham1995
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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

    • one year ago
  28. shakir
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    ok then wat abt range

    • one year ago
  29. heena
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    but gautham dont u add lift's velocity too??

    • one year ago
  30. goutham1995
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    we dont have to ..think and tell

    • one year ago
  31. shakir
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    But not getting!!!!

    • one year ago
  32. shakir
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    Range should be 0.404 m somehow///

    • one year ago
  33. heena
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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?

    • one year ago
  34. goutham1995
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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?

    • one year ago
  35. heena
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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

    • one year ago
  36. shakir
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    Any Idea Guys!

    • one year ago
  37. goutham1995
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    well i dont know how to do it in heena's way, although we can

    • one year ago
  38. shakir
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    wat abt Range!!!!!!!!!

    • one year ago
  39. goutham1995
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    im getting 0.044

    • one year ago
  40. heena
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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

    • one year ago
  41. goutham1995
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    so either time of flight isnt correct or shakir didnt type properly

    • one year ago
  42. shakir
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    no i am correct!

    • one year ago
  43. heena
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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

    • one year ago
  44. shakir
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    Confused!!

    • one year ago
  45. shakir
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    Time Wasting lets move to another question

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

    • one year ago
  47. goutham1995
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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

    • one year ago
  48. phi
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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

    • one year ago
  49. goutham1995
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    @ heena wont it be a-g?

    • one year ago
  50. goutham1995
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    @heena *

    • one year ago
  51. phi
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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?

    • one year ago
  52. heena
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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

    • one year ago
  53. shivam_bhalla
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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

    • one year ago
  54. shivam_bhalla
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    Try the second part :P

    • one year ago
  55. shakir
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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

    • one year ago
  56. shakir
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    @shivam_bhalla

    • one year ago
  57. heena
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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?

    • one year ago
  58. shivam_bhalla
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    @heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s

    • one year ago
  59. heena
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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

    • one year ago
  60. goutham1995
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    @heena - he has resolved the components

    • one year ago
  61. heena
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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 :)

    • one year ago
  62. shivam_bhalla
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    @heena , You should not take the acceleration of lift into account. Think about it.

    • one year ago
  63. heena
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    but y not @shivam_bhalla lift is not static u cant neglect that :O

    • one year ago
  64. shakir
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    @shivam_bhalla WAT ABT RANGE???

    • one year ago
  65. shivam_bhalla
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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

    • one year ago
  66. shakir
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    PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

    • one year ago
  67. shivam_bhalla
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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

    • one year ago
  68. shakir
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    YES THEN THE DISTANCE IS ???????

    • one year ago
  69. shakir
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    @shivam_bhalla

    • one year ago
  70. heena
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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

    • one year ago
  71. shivam_bhalla
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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

    • one year ago
  72. shakir
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    HOW U=0.2?

    • one year ago
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