Answer
Please see the work below.
Work Step by Step
(a) We know that
The position of the particle in the x direction is
$x=x_{\circ}+v_{\circ}t+(\frac{1}{2})a_xt^2$
$x=0+11t+(\frac{1}{2})(-1.2)t^2$
$x=11t-0.6t^2$
When the particle crosses the y-axis then x value is zero
$0=11t-0.6t^2$
This simplifies to :
$t=18s$
(b) The position of the particle in y direction is
$y=y_{\circ}+v_{\circ}t+(\frac{1}{2})a_yt^2$
We plug in the known values to obtain:
$y=0+145+(-\frac{1}{2})(0.26)t^2$
$y=14t+0.13t^2$
At $t=18s$
$y=14(18)+0.13(18)^2$
$y=290m$
