## mathslover 4 years ago A particle is projected with speed $$\sqrt{3} i\cap + j\cap +k\cap$$ from the origin. If the highest point of trajectory , the velocity vector makes and angle $$\theta$$ with n-axis, then find $$\tan \theta$$

1. mathslover

I don't know how to represent i cap

2. mathslover

@sauravshakya

3. mathslover

|dw:1349108807115:dw|

4. anonymous

Three Co-orDinates...hm......if it is Two Cordinate tanx = vy/vx @highest point of trajectory vertical component of velocity is 0

5. mathslover

$u\sin \theta = 0$

6. mathslover

Now? sin theta = 0?

7. anonymous

if this is Two-Cordinate then usinx =0........Since it is a Three Coordinate...) i dont thhink...so...and i have nt studied....Three Coordinate Plane...

8. mathslover

hmn no problem. @experimentX @sauravshakya @helder_edwin any ideas?

9. mathslover

@Jemurray3 @mukushla @ghazi @ganeshie8

10. Ghazi

@mathslover what does n axis refer to? this is supposed to be a three dimensional coordinate system

11. Ghazi

and theta= 30 degrees

12. mathslover

Well the question dont tells this .... I am also confused , also in starting it is written : Treat vertically upward as K|dw:1349109611966:dw|

13. Ghazi

i mean $\tan \theta = \frac{ 1 }{ \sqrt3 }$

14. mathslover

You're correct @ghazi but HOW?

15. Ghazi

see if you consider a three dimensional system check out the spherical coordinate system on google you'll find theta is the angle between x and y axis now $\tan \theta= \frac{ y }{ x }=\frac{ 1 }{ \sqrt3 }$ x and y are the intercepts

16. mathslover

Oh very easy it was, I feel it now. Thanks a ton @ghazi

17. Ghazi

:)

18. amistre64

this is what i was thinking |dw:1349110013381:dw|

19. Ghazi

@amistre64 nice work, @mathslover this is what i was talking about .....