Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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)
 one year ago
 one year 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)
 one year ago
 one year ago

This Question is Closed

shakirBest ResponseYou've already chosen the best response.0
@yash2651995 plzz help!
 one year ago

shakirBest ResponseYou've already chosen the best response.0
@yash2651995 i posted this question already but did nt understand make u make me understnad
 one year ago

yash2651995Best 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 :)
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
I somehow got the first part of question's realtion. but final part not getting.
 one year ago

yash2651995Best 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
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
Sorry, \[ H_r = 2ut\cos \theta \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
v = u+at \[ t = u\sin\theta/g\]
 one year ago

heenaBest ResponseYou've already chosen the best response.3
the qn is done or stilll need help @shakir
 one year ago

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

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

experimentXBest 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
 one year ago

experimentXBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection
 one year ago

shakirBest ResponseYou've already chosen the best response.0
The answer should be degree
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
Wat's the final answer?
 one year ago

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

yash2651995Best ResponseYou've already chosen the best response.0
neh neh, that was some other question's answer.. my workbook is a mess..
 one year ago

experimentXBest 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 ...
 one year ago

experimentXBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i'll check and reply back
 one year ago

experimentXBest 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
 one year ago

heenaBest 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
 one year ago

heenaBest 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
 one year ago

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

experimentXBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
mathematics seems a lot simpler than physics!!
 one year ago

heenaBest ResponseYou've already chosen the best response.3
it might be but i m not a studnt of maths so dunno much abu dat :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.