A particle is thrown with speed `u` at an angle `theta` with horizontal.

Find the average velocity of particle:
(a) `O` to `C` (b) `A` to `B` ( c) `O` to `H`

Explanation

(a) `O` to `C` : Displacement in vertical direction `= 0`, hence `bar(v)_(y) = 0`.
In the horizontal direction, velocity is constant, hence
`bar(v)_(x) = u cos theta`
`bar(v) = bar(v)_(x) = u cos theta`
(b) `A` to `B`: Since the vertical distance between `A` and `B = 0` ltbgt `bar(v)_(y) = 0 , bar(v)_(x) = u cos theta, bar(v) = u cos theta`
( c) `O` to `H`: `bar(v)_(x) = u cos theta`
`bar(v)_(y) = (H_(max))/((T)/(2)) = (u sin theta)/(2)`
or `bar(v)_(y) = (u sin theta + 0)/(2) = (u sin theta)/(2)`
`v = sqrt(bar(v)_(x)^(2) + bar(v)_(y)^(2)) = (u)/(2)sqrt(1 + 3 cos^(2) theta)`