A particle is moving in x-y plane. Its initial velocity and acceleration are u=(4i^+8j^)m/s and a=(2i^-4j^)m/s2. 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, Sy=uyt+12ayt2 -10=8t-12×4×t2 By solving this equation we will get t = 5 s