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

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.0i'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 :)

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

yash2651995
 2 years ago
Best 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

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1So we have \[ (u \sin\theta)^2/2g = 2u\cos\theta u \sin\theta/g\] \[ (u \sin\theta)^ = 4u^2\cos\theta \sin\theta\] \[ \sin\theta = 4 \cos\theta \]

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

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

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1now 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
 2 years ago
Best 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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://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

heena
 2 years ago
Best ResponseYou've already chosen the best response.3that 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

heena
 2 years ago
Best ResponseYou've already chosen the best response.3or 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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1All 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

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

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