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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
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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shakir Group TitleBest ResponseYou've already chosen the best response.0
@yash2651995 plzz help!
 2 years ago

shakir Group TitleBest ResponseYou've already chosen the best response.0
@yash2651995 i posted this question already but did nt understand make u make me understnad
 2 years ago

yash2651995 Group TitleBest ResponseYou've already chosen the best response.0
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

shakir Group TitleBest ResponseYou've already chosen the best response.0
OK Plz go on!
 2 years ago

heena Group TitleBest ResponseYou've already chosen the best response.3
yea yash go on :)
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
I somehow got the first part of question's realtion. but final part not getting.
 2 years ago

yash2651995 Group TitleBest ResponseYou've already chosen the best response.0
@heena nhi aya toh tum hi samjha diyo!! after all you are an excellent teacher !! :P :D
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[V_r = (u\sin\theta)^2/2g\] \[H_r = u t\cos \theta\] Given that \[ 2V_r = H_r \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Sorry, \[ H_r = 2ut\cos \theta \]
 2 years ago

shakir Group TitleBest ResponseYou've already chosen the best response.0
Hr is Range??
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
v = u+at \[ t = u\sin\theta/g\]
 2 years ago

heena Group TitleBest ResponseYou've already chosen the best response.3
the qn is done or stilll need help @shakir
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

heena Group TitleBest ResponseYou've already chosen the best response.3
well @shakir @experimentX is helping u try to cooperate with him ok :) u ll undrstnd easily :)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection
 2 years ago

shakir Group TitleBest ResponseYou've already chosen the best response.0
The answer should be degree
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Wat's the final answer?
 2 years ago

heena Group TitleBest ResponseYou've already chosen the best response.3
i guess this is already solved @shakir http://openstudy.com/study#/updates/4fc4c3d8e4b0964abc8718ab
 2 years ago

yash2651995 Group TitleBest ResponseYou've already chosen the best response.0
neh neh, that was some other question's answer.. my workbook is a mess..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ t = \frac x{u \cos(75.96)}\] 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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i'll check and reply back
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

heena Group TitleBest ResponseYou've already chosen the best response.3
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

heena Group TitleBest ResponseYou've already chosen the best response.3
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

heena Group TitleBest ResponseYou've already chosen the best response.3
srry for wrong statement its here U=initial velcoity R=u^2/g
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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%3D4a2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc find the slope at R/4 http://www.wolframalpha.com/input/?i=solve+0%3D4a2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc take tan inverse which is pi/4
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
mathematics seems a lot simpler than physics!!
 2 years ago

heena Group TitleBest ResponseYou've already chosen the best response.3
it might be but i m not a studnt of maths so dunno much abu dat :)
 2 years ago
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