The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$

$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).

$A$ projectile is fired from level ground at an angle $\theta$ above the horizontal. The elevation angle $\phi$ of the highest point as seen from the launch point is related to $\theta$ by the relation

If the time of flight of a bullet over a horizontal range $R$ is $T \ s$,the inclination of the direction of projection to the horizontal is

If bullets are fired in all possible directions from the same point with an equal velocity of $10 \,m \,s^{-1}$ and an angle of projection $45^{\circ}$,then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,m^2$)

$A$ cricket player hits a ball like a projectile,and the fielder catches the ball after $2 \ s$. The maximum height reached by the ball is $(g = 10 \ m/s^2)$. (in $m$)