A particle starts from origin at `t = 0` with a velocity of `15 hati ms^(-1)` and moves in xy-plane under the action of a force which produces a constant acceleration of `15 hati + 20 hatj ms^(-2)`. Find the y-coordinate of the particle at the instant its x-coordinate is 180 m.

Explanation

The position of the particle is given by
`r(t) = v_(0)t +(1)/(2) a t^(2) = 15 hati t +(1)/(5) (15 hati + 20 hatj) t^(2)`
`= (15 t + 7.5 t^(2)) hati + 10 hatj t^(2)`
`:. X(t) = 15 t +7.5 t^(2)` and `y(t) = 10 t^(2)`
If `x(t) = 180m,t = ?`
`180 = 15 t +7.5 t^(2) rArr t = 4s`
`:.` y-coordinate, `y(t) = 10 xx 16 = 160 m`