## 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)

1. shakir

@yash2651995 plzz help!

2. shakir

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

3. 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 :)

4. shakir

OK Plz go on!

5. heena

yea yash go on :)

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

7. yash2651995

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

8. experimentX

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

9. shakir

ok

10. experimentX

Sorry, $H_r = 2ut\cos \theta$

11. shakir

Hr is Range??

12. experimentX

v = u+at $t = u\sin\theta/g$

13. heena

the qn is done or stilll need help @shakir

14. shakir

i need help

15. experimentX

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

16. heena

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

17. shakir

ok

18. 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

19. experimentX
20. experimentX

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

21. shakir

22. shakir

45 degree

24. heena

25. yash2651995

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

26. 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 ...

27. 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

28. experimentX

29. experimentX
30. heena

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

31. heena

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

32. heena

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

33. 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

34. experimentX

mathematics seems a lot simpler than physics!!

35. heena

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