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shakir Group Title

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)

  • 2 years ago
  • 2 years ago

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  1. shakir Group Title
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    @yash2651995 plzz help!

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

    • 2 years ago
  3. yash2651995 Group Title
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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 :)

    • 2 years ago
  4. shakir Group Title
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    OK Plz go on!

    • 2 years ago
  5. heena Group Title
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    yea yash go on :)

    • 2 years ago
  6. Aadarsh Group Title
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    I somehow got the first part of question's realtion. but final part not getting.

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

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

    • 2 years ago
  9. shakir Group Title
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    ok

    • 2 years ago
  10. experimentX Group Title
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    Sorry, \[ H_r = 2ut\cos \theta \]

    • 2 years ago
  11. shakir Group Title
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    Hr is Range??

    • 2 years ago
  12. experimentX Group Title
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    v = u+at \[ t = u\sin\theta/g\]

    • 2 years ago
  13. heena Group Title
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    the qn is done or stilll need help @shakir

    • 2 years ago
  14. shakir Group Title
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    i need help

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

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

    • 2 years ago
  17. shakir Group Title
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    ok

    • 2 years ago
  18. experimentX Group Title
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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

    • 2 years ago
  19. experimentX Group Title
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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

    • 2 years ago
  20. experimentX Group Title
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    so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection

    • 2 years ago
  21. shakir Group Title
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    The answer should be degree

    • 2 years ago
  22. shakir Group Title
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    45 degree

    • 2 years ago
  23. Aadarsh Group Title
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    Wat's the final answer?

    • 2 years ago
  24. heena Group Title
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    i guess this is already solved @shakir http://openstudy.com/study#/updates/4fc4c3d8e4b0964abc8718ab

    • 2 years ago
  25. yash2651995 Group Title
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    neh neh, that was some other question's answer.. my workbook is a mess..

    • 2 years ago
  26. experimentX Group Title
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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 ...

    • 2 years ago
  27. experimentX Group Title
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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

    • 2 years ago
  28. experimentX Group Title
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    i'll check and reply back

    • 2 years ago
  29. heena Group Title
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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

    • 2 years ago
  30. heena Group Title
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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

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

    • 2 years ago
  32. experimentX Group Title
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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

    • 2 years ago
  33. experimentX Group Title
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    mathematics seems a lot simpler than physics!!

    • 2 years ago
  34. heena Group Title
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    it might be but i m not a studnt of maths so dunno much abu dat :)

    • 2 years ago
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