If a body projected with a velocity of $19.6 \ m/s$ reaches a maximum height of $9.8 \ m$, then the range of the projectile is (Neglect air resistance). (in $m$)

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........

$A$ ball is thrown at different angles with the same speed $u$ and from the same point. It has the same range in both cases. If $y_1$ and $y_2$ are the heights attained in the two cases,then $y_1 + y_2$ is equal to: