The vertical displacement ( $y$ in metre) of a projectile in term of its horizontal displacement ( $x$ in metre) is given by $y=\left(\sqrt{3} x-0.2 x^2\right)$. The time of flight of the projectile is

(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A body projected at certain angle $\left(\neq 90^{\circ}\right)$ from the ground crosses a point in its path at a time of 2.3 s and from there it reaches the ground after a time of 5.7 s . The maximum heigh reached by the body is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times 2 s and 8 s respectively. The horizontal distance between the two points is

(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

If a body projected with a velocity of $19.6 \mathrm{~ms}^{-1}$ reaches a maximum height of 9.8 m , then the range of the projectile is

(Neglect air resistance)