At `t = 0`, the particle is at the origin and its velocity is `10 m//s` at an angle `37^(@)` with the x-axis. The particle moves in `x-y` plane with a constant acceleration of `1 m//s^(2)` along the y-axis. Find the magnitude and direction of the velocity after `2 s`.

Explanation

First, resolve the velocity along the coordinate axes


`x` : (uniform acceleration): `= v_(x) = u_(x) = 8 m//s`
`y` : (constant acceleration) : `v_(y) = u_(y) + a_(y)t = 6 + 1 xx 2 = 8 m//s`
Resultant velocity `v = sqrt(v_(x)^(2) +v_(y)^(2)) = 8sqrt(2) m//s`
`tan theta = (v_(y))/(v_(x)) = 1 rArr = 45^(@)`