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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)

Physics
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@yash2651995 plzz help!
@yash2651995 i posted this question already but did nt understand make u make me understnad
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 :)

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Other answers:

OK Plz go on!
yea yash go on :)
I somehow got the first part of question's realtion. but final part not getting.
@heena nhi aya toh tum hi samjha diyo!! after all you are an excellent teacher !! :P :D
\[V_r = (u\sin\theta)^2/2g\] \[H_r = u t\cos \theta\] Given that \[ 2V_r = H_r \]
ok
Sorry, \[ H_r = 2ut\cos \theta \]
Hr is Range??
v = u+at \[ t = u\sin\theta/g\]
the qn is done or stilll need help @shakir
i need help
So we have \[ (u \sin\theta)^2/2g = 2u\cos\theta u \sin\theta/g\] \[ (u \sin\theta)^ = 4u^2\cos\theta \sin\theta\] \[ \sin\theta = 4 \cos\theta \]
well @shakir @experimentX is helping u try to co-operate with him ok :) u ll undrstnd easily :)
ok
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
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
so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection
The answer should be degree
45 degree
Wat's the final answer?
neh neh, that was some other question's answer.. my workbook is a mess..
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 ...
\[ t = \frac x{u \cos(75.96)}\] 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
i'll check and reply back
http://www.wolframalpha.com/input/?i=d%2Fdx+%2825*sin%281.3258%29%28x%2F%282*25*cos%281.3258%29%29%29+-+1%2F2*9.8*%28x%2F%282*25*cos%281.3258%29%29%29^2%29+%2C+x%3D60%2F4 Ah it's around 45 degrees
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
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
srry for wrong statement its here U=initial velcoity R=u^2/g
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
mathematics seems a lot simpler than physics!!
it might be but i m not a studnt of maths so dunno much abu dat :)

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