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shakir

  • 3 years ago

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?

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  1. ArchiePhysics
    • 3 years ago
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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. ArchiePhysics
    • 3 years ago
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    it then falls and its motion is described by another equation

  3. ArchiePhysics
    • 3 years ago
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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)

  4. shakir
    • 3 years ago
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    @ArchiePhysics so wat is ur answer!

  5. ArchiePhysics
    • 3 years ago
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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

  6. ArchiePhysics
    • 3 years ago
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    Do you have any questions?

  7. shakir
    • 3 years ago
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    |dw:1338391009065:dw|

  8. shakir
    • 3 years ago
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    @ArchiePhysics

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

  10. shakir
    • 3 years ago
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    it s tan alpha

  11. shakir
    • 3 years ago
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    it is the answer

  12. shakir
    • 3 years ago
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    |dw:1338391169913:dw|

  13. yash2651995
    • 3 years ago
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    s tan theta

  14. shakir
    • 3 years ago
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    yes

  15. yash2651995
    • 3 years ago
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    wait.. full explanation is on the way..

  16. shakir
    • 3 years ago
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    it is the answer @yash2651995

  17. shakir
    • 3 years ago
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    @yash2651995 is that expl frm u??

  18. yash2651995
    • 3 years ago
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    uploading my solution

  19. shakir
    • 3 years ago
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    ok waiting

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

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

  22. shakir
    • 3 years ago
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    @yash2651995 thanzzz

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

  24. yash2651995
    • 3 years ago
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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

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

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

  27. yash2651995
    • 3 years ago
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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 ♪

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

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