Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

shakir

  • 2 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?

  • This Question is Closed
  1. ArchiePhysics
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it then falls and its motion is described by another equation

  3. ArchiePhysics
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ArchiePhysics so wat is ur answer!

  5. ArchiePhysics
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you have any questions?

  7. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1338391009065:dw|

  8. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ArchiePhysics

  9. ArchiePhysics
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Didn't understand, absolutely. What is that?

  10. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it s tan alpha

  11. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is the answer

  12. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1338391169913:dw|

  13. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    s tan theta

  14. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  15. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    wait.. full explanation is on the way..

  16. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is the answer @yash2651995

  17. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @yash2651995 is that expl frm u??

  18. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    uploading my solution

  19. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok waiting

  20. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    compressed picture. original is taking lots of time.. i'd still upload it later

    1 Attachment
  21. Aadarsh
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @yash2651995 bhai, thanks for the answer. Can u explain how?|dw:1338391988800:dw|

  22. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @yash2651995 thanzzz

  23. shakir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanzz a billion @yash2651995 r u frm DElhi!!!!

  24. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @shakir neh yarr i'm from U.P.; a small suburb near Vns.

  26. Aadarsh
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na

  27. yash2651995
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........

  29. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.