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A particle is projected with velocity u ...

A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

Text Solution

Verified by Experts

The correct Answer is:
`(4)`
Initial velocity
`u = u cos theta hat i+ u sin theta hat j`
Velocity after time `t` is
`v = u cos theta hat i+ (u sin theta - "gt") hat j`
Since `u and v` are perpendicular to each other, therefore,
`vec u. vec v = 0`
`(u cos theta hat i+ u sin theta hat j) xx [u cos theta hat i+(u sin theta - gt) hat j] = 0`
`u^2 cos^2 theta + u^2 sin^2 theta - u sin theta "gt" = 0`
`t = (u)/(g sin theta) = (20)/(10 xx (1)/(2)) = 4 s`.
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Knowledge Check

  • A particle is projected with a velocity u making an angle theta with the horizontal. At any instant its velocity becomes v which is perpendicular to the initial velocity u. Then v is

    A
    `u cos theta`
    B
    `u tan theta`
    C
    `u sec theta`
    D
    `ucot theta`

  • A particle is thrown with a speed u at an angle theta with horizontal. After how much time the velocity of particle will be perpendicular to the initial motion of direction

    A
    `(u)/(g cos theta)`
    B
    `(u)/(g sin theta)`
    C
    `(u sin theta)/(g)`
    D
    `(u cos theta)/(g)`

  • A particle is projected with velocity v at an angle of theta with horizontal. The average angular velocity of the particle from the point of projection to impact equals:

    A
    `(g cos theta)/(thetav)`
    B
    `g/(v sin theta)`
    C
    `g/(v theta)`
    D
    `(g theta)/(v sin theta)`