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shakir

A particle is projected upwards from level ground at an angle with horizontal such that its range is twice the maximum height reached by it.The angle made by its velocity vector with the horizontal at ahorizontal distance R/4 from the point of projection(R is horizontal range)

  • one year ago
  • one year ago

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  1. shakir
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    @yash2651995 plzz help!

    • one year ago
  2. shakir
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    @yash2651995 i posted this question already but did nt understand make u make me understnad

    • one year ago
  3. yash2651995
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    i'll do it in a systematic way.. BTW heena's way was right she found a connection.. and took out the answer.. if anyone else can also do it, it'd be appreciated.. as there are lots of ways to reach the answer specially in kinematics :)

    • one year ago
  4. shakir
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    OK Plz go on!

    • one year ago
  5. heena
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    yea yash go on :)

    • one year ago
  6. Aadarsh
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    I somehow got the first part of question's realtion. but final part not getting.

    • one year ago
  7. yash2651995
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    @heena nhi aya toh tum hi samjha diyo!! after all you are an excellent teacher !! :P :D

    • one year ago
  8. experimentX
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    \[V_r = (u\sin\theta)^2/2g\] \[H_r = u t\cos \theta\] Given that \[ 2V_r = H_r \]

    • one year ago
  9. shakir
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    ok

    • one year ago
  10. experimentX
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    Sorry, \[ H_r = 2ut\cos \theta \]

    • one year ago
  11. shakir
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    Hr is Range??

    • one year ago
  12. experimentX
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    v = u+at \[ t = u\sin\theta/g\]

    • one year ago
  13. heena
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    the qn is done or stilll need help @shakir

    • one year ago
  14. shakir
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    i need help

    • one year ago
  15. experimentX
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    So we have \[ (u \sin\theta)^2/2g = 2u\cos\theta u \sin\theta/g\] \[ (u \sin\theta)^ = 4u^2\cos\theta \sin\theta\] \[ \sin\theta = 4 \cos\theta \]

    • one year ago
  16. heena
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    well @shakir @experimentX is helping u try to co-operate with him ok :) u ll undrstnd easily :)

    • one year ago
  17. shakir
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    ok

    • one year ago
  18. experimentX
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    lol ... i am not sure ... check this http://www.wolframalpha.com/input/?i=parametric+plot+x%3D2*30*cos%281.3258%29t+%2C+y+%3D+30*sin%281.3258%29t+-+1%2F2*9.8*t^2

    • one year ago
  19. experimentX
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    I guess it worked http://www.wolframalpha.com/input/?i=parametric+plot+x%3D2*25*cos%281.3258%29t+%2C+y+%3D+25*sin%281.3258%29t+-+1%2F2*9.8*t^2

    • one year ago
  20. experimentX
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    so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection

    • one year ago
  21. shakir
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    The answer should be degree

    • one year ago
  22. shakir
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    45 degree

    • one year ago
  23. Aadarsh
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    Wat's the final answer?

    • one year ago
  24. heena
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    i guess this is already solved @shakir http://openstudy.com/study#/updates/4fc4c3d8e4b0964abc8718ab

    • one year ago
  25. yash2651995
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    neh neh, that was some other question's answer.. my workbook is a mess..

    • one year ago
  26. experimentX
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    now you have relation ... \[ y(x) = u t \sin(75.96) - \frac 12 gt^2 , \; \; x(t) = u\cos(75.96) t\] The horizontal range is given by relation \[ H_r = 2u^2\cos(75.96)\sin(75.96)/g\] Find, dy/dx = dy/dt * 1/(dx/dt)| H_r/4 tan inverse of this value should be your answer ...

    • one year ago
  27. experimentX
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    \[ t = \frac x{u \cos(75.96)}\] So we have \[ y = x \tan(75.96) - \frac {g x^2}{2 u^2 \cos^2(75.96) } \] \[ \frac{dy}{dx} = \tan(75.96) - \frac {g x}{u^2 \cos^2(75.96) }, x = Hr/4=2u^2\cos(75.96)\sin(75.96)/4g = \\ u^2\cos(75.96)\sin(75.96)/2g\] \[ =\tan(75.96) - \frac{gu^2\cos(75.96)\sin(75.96)/2g}{u^2 \cos^2(75.96)} \] \[ =\tan(75.96) - \frac{\sin(75.96)/2}{\cos(75.96)} \] Looks like i didn't get 45

    • one year ago
  28. experimentX
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    i'll check and reply back

    • one year ago
  29. heena
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    that too lengthy method @experimentX we can also use it like this way R=2hmax if R get 1/4 means 2Hmax will also get 1/4 R/4=Hmax/2 we know Hmax=u^2/2g R/4=u^2/4g [R=u^2sin2theta/g and Hmax=u^2/2g] u^2sin2theta/4g=u^2/4g sin2theta=1 2theta=90 theta=90/2 theta=45

    • one year ago
  30. heena
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    or there is one more way R=2hmax if R get 1/4 means 2Hmax will also get 1/4 R/4=Hmax/2 we know Hmax=u^2/2g R/4=u^2/4g here U=initial velcoity R=u^2/2g this is the formula of max range and it comes in 45degree

    • one year ago
  31. heena
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    srry for wrong statement its here U=initial velcoity R=u^2/g

    • one year ago
  32. experimentX
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    All messed up in equation ... :( best way is to model a parabola y = ax^2 + bx + c put some points (-2,0), (0,2), (2,0) solve for a,b,c http://www.wolframalpha.com/input/?i=solve+0%3D4a-2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc find the slope at R/4 http://www.wolframalpha.com/input/?i=solve+0%3D4a-2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc take tan inverse which is pi/4

    • one year ago
  33. experimentX
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    mathematics seems a lot simpler than physics!!

    • one year ago
  34. heena
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    it might be but i m not a studnt of maths so dunno much abu dat :)

    • one year ago
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