## 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

1. shakir Group Title

The answer is (a) 0.2s (b) 0.404m

2. apoorvk Group Title

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 Group Title

Since it was thrown with an angle why we need to consider the acceleration

4. shakir Group Title

T=2usintheta/g can we use that??

5. heena Group Title

hey is time taken =1sec

6. heena Group Title

sorry i mean 2sec ??

7. heena Group Title

and yea shakir u can use dat formula u mention above

8. shakir Group Title

How in question it is given that theta=30

9. apoorvk Group Title

10. shakir Group Title

@heena i got the Time of Flight Plzz help me to do distance

11. heena Group Title

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 Group Title

so wats the time of flight am i right?? i mean my ans is correct??

13. shakir Group Title

0.2s

14. heena Group Title

but how?

15. shakir Group Title

i am also confused i did it with out taking any relative acceleration and velocity

16. heena Group Title

still show me ur work wat u did

17. shakir Group Title

|dw:1338466097736:dw|

18. heena Group Title

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 Group Title

cany=*can

20. heena Group Title

and yea shakir for the second option u have to cal range means u^2sin2theta/g

21. shakir Group Title

not getting that

22. heena Group Title

i the distance 0.2sqrt3 ??

23. heena Group Title

do u knw sin2theta means 2xsinthetaxcostheta

24. shakir Group Title

yes

25. shakir Group Title

|dw:1338466702370:dw|

26. shakir Group Title

but that is wrong

27. goutham1995 Group Title

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 Group Title

ok then wat abt range

29. heena Group Title

but gautham dont u add lift's velocity too??

30. goutham1995 Group Title

we dont have to ..think and tell

31. shakir Group Title

But not getting!!!!

32. shakir Group Title

Range should be 0.404 m somehow///

33. heena Group Title

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 Group Title

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 Group Title

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 Group Title

Any Idea Guys!

37. goutham1995 Group Title

well i dont know how to do it in heena's way, although we can

38. shakir Group Title

wat abt Range!!!!!!!!!

39. goutham1995 Group Title

im getting 0.044

40. heena Group Title

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 Group Title

so either time of flight isnt correct or shakir didnt type properly

42. shakir Group Title

no i am correct!

43. heena Group Title

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 Group Title

Confused!!

45. shakir Group Title

Time Wasting lets move to another question

46. heena Group Title

ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a

47. goutham1995 Group Title

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 Group Title

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 Group Title

@ heena wont it be a-g?

50. goutham1995 Group Title

@heena *

51. phi Group Title

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 Group Title

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 Group Title

@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 Group Title

Try the second part :P

55. shakir Group Title

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 Group Title

@shivam_bhalla

57. heena Group Title

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 Group Title

@heena , mistake 1 = assume g = 10 m/s mistake 2 = u = 2 m/s

59. heena Group Title

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 Group Title

@heena - he has resolved the components

61. heena Group Title

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 Group Title

@heena , You should not take the acceleration of lift into account. Think about it.

63. heena Group Title

but y not @shivam_bhalla lift is not static u cant neglect that :O

64. shakir Group Title

@shivam_bhalla WAT ABT RANGE???

65. shivam_bhalla Group Title

@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 Group Title

PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

67. shivam_bhalla Group Title

@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 Group Title

YES THEN THE DISTANCE IS ???????

69. shakir Group Title

@shivam_bhalla

70. heena Group Title

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 Group Title

@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 Group Title

HOW U=0.2?