A particle is projected downwards from point 'A' with initial velocity $5m/s$ at an angle $\theta ={37}^{o}C$ with the vertical. A strong horizontal wind gives the particle a constant horizontal acceleration of $6m/s^2$ in the x - direction. If the particle strikes the ground directly under its released position, the height 'h' of point A (in $m$) is (answer upto two decimal places). The downward acceleration is taken as $g=10m/s^2$ , take (sin37°=3/5, cos37°=4/5)