An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The horizontal range of the particle is equal to

$A$ ball thrown by one player reaches another player in $2 \text{ s}$. The maximum height attained by the ball above the point of projection will be $.... \text{ m}$.

Match the columns:

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

$A$ particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$, then the angle of projection to hit the target is

$A$ player kicks a football with an initial speed of $25\, ms^{-1}$ at an angle of $45^{\circ}$ from the ground. What are the maximum height and the time taken by the football to reach the highest point during motion? (Take $g = 10\, ms^{-2}$)

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is