A particle moves in the `x-y` plane such that its coordinates are given `x = 10sqrt(3) t, y = 10t - t^(2)` and `t` is time, find the initial velocity of the particle.

Explanation

`x = 10sqrt(3) t`
`v_(x) = (dx)/(d t) = 10 sqrt(3)`
`y = 10t - t^(2)`
`v_(y) = (dy)/(dt) = 10 - 2t`
At `t = 0, v_(x) = 10sqrt(3) m//s`
`v_(y) = 10 - 2t = 10 - 2 xx 0 = 10 m//s`
Initial velocity `v = sqrt(v_(x)^(2) + v_(y)^(2)) = 20 m//s`
`tan theta = (v_(y))/(v_(x)) = (10)/(10sqrt(3)) = (1)/(sqrt(3)) rArr theta = 30^(@)`