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shakir Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. ArchiePhysics Group Title
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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.

    • 2 years ago
  2. ArchiePhysics Group Title
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    it then falls and its motion is described by another equation

    • 2 years ago
  3. ArchiePhysics Group Title
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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)

    • 2 years ago
  4. shakir Group Title
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    @ArchiePhysics so wat is ur answer!

    • 2 years ago
  5. ArchiePhysics Group Title
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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

    • 2 years ago
  6. ArchiePhysics Group Title
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    Do you have any questions?

    • 2 years ago
  7. shakir Group Title
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    |dw:1338391009065:dw|

    • 2 years ago
  8. shakir Group Title
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    @ArchiePhysics

    • 2 years ago
  9. ArchiePhysics Group Title
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    Didn't understand, absolutely. What is that?

    • 2 years ago
  10. shakir Group Title
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    it s tan alpha

    • 2 years ago
  11. shakir Group Title
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    it is the answer

    • 2 years ago
  12. shakir Group Title
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    |dw:1338391169913:dw|

    • 2 years ago
  13. yash2651995 Group Title
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    s tan theta

    • 2 years ago
  14. shakir Group Title
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    yes

    • 2 years ago
  15. yash2651995 Group Title
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    wait.. full explanation is on the way..

    • 2 years ago
  16. shakir Group Title
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    it is the answer @yash2651995

    • 2 years ago
  17. shakir Group Title
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    @yash2651995 is that expl frm u??

    • 2 years ago
  18. yash2651995 Group Title
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    uploading my solution

    • 2 years ago
  19. shakir Group Title
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    ok waiting

    • 2 years ago
  20. yash2651995 Group Title
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    compressed picture. original is taking lots of time.. i'd still upload it later

    • 2 years ago
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  21. Aadarsh Group Title
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    @yash2651995 bhai, thanks for the answer. Can u explain how?|dw:1338391988800:dw|

    • 2 years ago
  22. shakir Group Title
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    @yash2651995 thanzzz

    • 2 years ago
  23. shakir Group Title
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    Thanzz a billion @yash2651995 r u frm DElhi!!!!

    • 2 years ago
  24. yash2651995 Group Title
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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

    • 2 years ago
  25. yash2651995 Group Title
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    @shakir neh yarr i'm from U.P.; a small suburb near Vns.

    • 2 years ago
  26. Aadarsh Group Title
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    Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na

    • 2 years ago
  27. yash2651995 Group Title
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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 ♪

    • 2 years ago
  28. Aadarsh Group Title
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    hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........

    • 2 years ago
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