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shakir

A Particle is projected upwards from ground with a speed u at an angle theta with the horizontal. At the same time another paricle is dropped from a height h if they collide at a horizontal distance s from the point of projection then h is?

  • one year ago
  • one year ago

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  1. ArchiePhysics
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    At the moment of collision t = s/v1x , where v1x - initial horizontal component of velocity of the first object for the second object y(t) = h - gt^2/2 = h - gs^2/(2 u^2(cos theta)^2), but this is true only if the collision occurs before the first object rises its maximum height.

    • one year ago
  2. ArchiePhysics
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    it then falls and its motion is described by another equation

    • one year ago
  3. ArchiePhysics
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    Sorry I got little confused.That's all ok, we are not considering vertical motion of the first object, so the resultant equation is true for every moment of time. At the moment of collision t = s/v1x , where v1x - initial horizontal component of velocity of the first object v1x = u cos theta for the second object y(t) = h - gt^2/2 = h - gs^2/(2 u^2(cos theta)^2)

    • one year ago
  4. shakir
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    @ArchiePhysics so wat is ur answer!

    • one year ago
  5. ArchiePhysics
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    The last equation is the answer y(t) = h - gs^2/(2 u^2(cos theta)^2) where y - is the height at which the second object is located h - is the initial height of it

    • one year ago
  6. ArchiePhysics
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    Do you have any questions?

    • one year ago
  7. shakir
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    |dw:1338391009065:dw|

    • one year ago
  8. shakir
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    @ArchiePhysics

    • one year ago
  9. ArchiePhysics
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    Didn't understand, absolutely. What is that?

    • one year ago
  10. shakir
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    it s tan alpha

    • one year ago
  11. shakir
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    it is the answer

    • one year ago
  12. shakir
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    |dw:1338391169913:dw|

    • one year ago
  13. yash2651995
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    s tan theta

    • one year ago
  14. shakir
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    yes

    • one year ago
  15. yash2651995
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    wait.. full explanation is on the way..

    • one year ago
  16. shakir
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    it is the answer @yash2651995

    • one year ago
  17. shakir
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    @yash2651995 is that expl frm u??

    • one year ago
  18. yash2651995
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    uploading my solution

    • one year ago
  19. shakir
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    ok waiting

    • one year ago
  20. yash2651995
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    compressed picture. original is taking lots of time.. i'd still upload it later

    • one year ago
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  21. Aadarsh
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    @yash2651995 bhai, thanks for the answer. Can u explain how?|dw:1338391988800:dw|

    • one year ago
  22. shakir
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    @yash2651995 thanzzz

    • one year ago
  23. shakir
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    Thanzz a billion @yash2651995 r u frm DElhi!!!!

    • one year ago
  24. yash2651995
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    i'm replacing theta with O v cosO is horizontal component of initial velocity of particle 'a'.. which will remain unaltered as no force thus acceleration acts on it (the horizontal component) hence, by eq, s= ut +1/2 at^2 a=0 u=u cosO so s= u cosO t

    • one year ago
  25. yash2651995
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    @shakir neh yarr i'm from U.P.; a small suburb near Vns.

    • one year ago
  26. Aadarsh
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    Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na

    • one year ago
  27. yash2651995
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    IIT ab toh teacher ke chamchon ke liye ho gaya hai :( ♪ ♫ apni toh naiyaa hai raam ke bharose ... apni bhi naiyya ko paar tu lagaai de ♪

    • one year ago
  28. Aadarsh
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    hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........

    • one year ago
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