When a player throws a ball,it reaches the other player in $4 \,s$. If the height of each player is $1.8 \,m$,the maximum height attained by the ball above the ground is (in $\,m$)

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

If a body $A$ of mass $M$ is thrown with velocity $v$ at an angle of $30^{\circ}$ to the horizontal and another body $B$ of the same mass is thrown with the same speed at an angle of $60^{\circ}$ to the horizontal,the ratio of the horizontal range of $A$ to $B$ will be:

The path of a projectile is given by the equation $y = ax - bx^2$,where $a$ and $b$ are constants,and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection,respectively. The maximum height attained by the projectile and the angle of projection are respectively:

The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t \text{ m}$ and $y = 8t - 5t^2 \text{ m}$. The range of the projectile in metres is:

Match the columns:

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$