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)

- anonymous

- jamiebookeater

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- anonymous

@yash2651995 plzz help!

- anonymous

@yash2651995 i posted this question already but did nt understand make u make me understnad

- yash2651995

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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## More answers

- anonymous

OK Plz go on!

- anonymous

yea yash go on :)

- anonymous

I somehow got the first part of question's realtion. but final part not getting.

- yash2651995

@heena nhi aya toh tum hi samjha diyo!!
after all you are an excellent teacher !! :P :D

- experimentX

\[V_r = (u\sin\theta)^2/2g\]
\[H_r = u t\cos \theta\]
Given that \[ 2V_r = H_r \]

- anonymous

ok

- experimentX

Sorry, \[ H_r = 2ut\cos \theta \]

- anonymous

Hr is Range??

- experimentX

v = u+at
\[ t = u\sin\theta/g\]

- anonymous

the qn is done or stilll need help @shakir

- anonymous

i need help

- experimentX

So we have
\[ (u \sin\theta)^2/2g = 2u\cos\theta u \sin\theta/g\]
\[ (u \sin\theta)^ = 4u^2\cos\theta \sin\theta\]
\[ \sin\theta = 4 \cos\theta \]

- anonymous

well @shakir @experimentX is helping u try to co-operate with him ok :) u ll undrstnd easily :)

- anonymous

ok

- experimentX

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

- experimentX

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

- experimentX

so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection

- anonymous

The answer should be degree

- anonymous

45 degree

- anonymous

Wat's the final answer?

- anonymous

i guess this is already solved @shakir
http://openstudy.com/study#/updates/4fc4c3d8e4b0964abc8718ab

- yash2651995

neh neh, that was some other question's answer.. my workbook is a mess..

- experimentX

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 ...

- experimentX

\[ t = \frac x{u \cos(75.96)}\]
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

- experimentX

i'll check and reply back

- experimentX

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

- anonymous

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

- anonymous

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

- anonymous

srry for wrong statement its
here U=initial velcoity R=u^2/g

- experimentX

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

- experimentX

mathematics seems a lot simpler than physics!!

- anonymous

it might be but i m not a studnt of maths so dunno much abu dat :)

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