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shakir

  • 2 years ago

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)

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  1. shakir
    • 2 years ago
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    @yash2651995 plzz help!

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

  3. yash2651995
    • 2 years ago
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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 :)

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

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

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

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

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

  9. shakir
    • 2 years ago
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    ok

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

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

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

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

  14. shakir
    • 2 years ago
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    i need help

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

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

  17. shakir
    • 2 years ago
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    ok

  18. experimentX
    • 2 years ago
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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

  19. experimentX
    • 2 years ago
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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

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

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

  22. shakir
    • 2 years ago
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    45 degree

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

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

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

  26. experimentX
    • 2 years ago
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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 ...

  27. experimentX
    • 2 years ago
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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

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

  29. heena
    • 2 years ago
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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

  30. heena
    • 2 years ago
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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

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

  32. experimentX
    • 2 years ago
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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

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

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

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