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shakir

  • 2 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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  1. shakir
    • 2 years ago
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    The answer is (a) 0.2s (b) 0.404m

  2. apoorvk
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

  11. heena
    • 2 years ago
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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

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

  13. shakir
    • 2 years ago
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    0.2s

  14. heena
    • 2 years ago
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    but how?

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

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

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

  18. heena
    • 2 years ago
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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??

  19. heena
    • 2 years ago
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    cany=*can

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

  21. shakir
    • 2 years ago
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    not getting that

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

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

  24. shakir
    • 2 years ago
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    yes

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

  26. shakir
    • 2 years ago
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    but that is wrong

  27. goutham1995
    • 2 years ago
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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

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

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

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

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

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

  33. heena
    • 2 years ago
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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?

  34. goutham1995
    • 2 years ago
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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?

  35. heena
    • 2 years ago
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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

  36. shakir
    • 2 years ago
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    Any Idea Guys!

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

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

  39. goutham1995
    • 2 years ago
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    im getting 0.044

  40. heena
    • 2 years ago
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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

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

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

  43. heena
    • 2 years ago
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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

  44. shakir
    • 2 years ago
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    Confused!!

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

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

  47. goutham1995
    • 2 years ago
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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

  48. phi
    • 2 years ago
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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

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

  50. goutham1995
    • 2 years ago
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    @heena *

  51. phi
    • 2 years ago
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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?

  52. heena
    • 2 years ago
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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

  53. shivam_bhalla
    • 2 years ago
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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

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

  55. shakir
    • 2 years ago
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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

  56. shakir
    • 2 years ago
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    @shivam_bhalla

  57. heena
    • 2 years ago
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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?

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

  59. heena
    • 2 years ago
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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

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

  61. heena
    • 2 years ago
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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 :)

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

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

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

  65. shivam_bhalla
    • 2 years ago
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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

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

  67. shivam_bhalla
    • 2 years ago
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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

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

  69. shakir
    • 2 years ago
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    @shivam_bhalla

  70. heena
    • 2 years ago
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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

  71. shivam_bhalla
    • 2 years ago
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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

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

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