A particle is moving in x-y plane. Its initial velocity and acceleration are $\mathbf{u}=(4\hat{\mathbf{i}}+8\hat{\mathbf{j}})m/s$ and $\mathbf{a}=(2\hat{\mathbf{i}}-4\hat{\mathbf{j}})m/s^2$. Find the time when the particle will cross the x-axis. Initial coordinates of particle are $(4m, 10m)$.  

Particle starts from point P. components of its initial velocity and acceleration are as shown in figure. 

At the time of crossing the x-axis, its y-coordinate should be zero ot its y-displacement (w.r.t initial point P) is -10 m.

Using the equation, $$\begin{gathered} S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ -10=8t-\frac{1}{2}\times 4\times t^2 \\ By solving this equation we will get t = 5 s \end{gathered}$$