## Essential University Physics: Volume 1 (4th Edition)

(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 $t=18s$ $y=14(18)+0.13(18)^2$ $y=290m$