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@yash2651995 plzz help!

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

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
\[ t = u\sin\theta/g\]

i need help

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

ok

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?

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

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

i'll check and reply back

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

mathematics seems a lot simpler than physics!!

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