mathslover
  • mathslover
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 \(\tan \theta\)
Physics
chestercat
  • chestercat
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mathslover
  • mathslover
I don't know how to represent i cap
mathslover
  • mathslover
mathslover
  • mathslover
|dw:1349108807115:dw|

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anonymous
  • anonymous
Three Co-orDinates...hm......if it is Two Cordinate tanx = vy/vx @highest point of trajectory vertical component of velocity is 0
mathslover
  • mathslover
\[u\sin \theta = 0\]
mathslover
  • mathslover
Now? sin theta = 0?
anonymous
  • 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...
mathslover
  • mathslover
hmn no problem. @experimentX @sauravshakya @helder_edwin any ideas?
mathslover
  • mathslover
ghazi
  • ghazi
@mathslover what does n axis refer to? this is supposed to be a three dimensional coordinate system
ghazi
  • ghazi
and theta= 30 degrees
mathslover
  • 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|
ghazi
  • ghazi
i mean \[\tan \theta = \frac{ 1 }{ \sqrt3 }\]
mathslover
  • mathslover
You're correct @ghazi but HOW?
ghazi
  • 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
mathslover
  • mathslover
Oh very easy it was, I feel it now. Thanks a ton @ghazi
ghazi
  • ghazi
:)
amistre64
  • amistre64
this is what i was thinking |dw:1349110013381:dw|
ghazi
  • ghazi
@amistre64 nice work, @mathslover this is what i was talking about .....

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