Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

I don't know how to represent i cap
|dw:1349108807115:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Three Co-orDinates...hm......if it is Two Cordinate tanx = vy/vx @highest point of trajectory vertical component of velocity is 0
\[u\sin \theta = 0\]
Now? sin theta = 0?
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...
hmn no problem. @experimentX @sauravshakya @helder_edwin any ideas?
@mathslover what does n axis refer to? this is supposed to be a three dimensional coordinate system
and theta= 30 degrees
Well the question dont tells this .... I am also confused , also in starting it is written : Treat vertically upward as K|dw:1349109611966:dw|
i mean \[\tan \theta = \frac{ 1 }{ \sqrt3 }\]
You're correct @ghazi but HOW?
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
Oh very easy it was, I feel it now. Thanks a ton @ghazi
:)
this is what i was thinking |dw:1349110013381:dw|
@amistre64 nice work, @mathslover this is what i was talking about .....

Not the answer you are looking for?

Search for more explanations.

Ask your own question