For a projectile,the ratio of maximum height reached to the square of flight time is $(g = 10 \ m/s^2)$.

The maximum height reached by a projectile is $64 \,m$. If the initial velocity is halved, the new maximum height of the projectile is . . . . . . $m$.

$A$ projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with the vertical and the range is $R_2$. Then:

For a projectile, if $\alpha$ is the angle of projection, $R$ is the range, $h$ is the maximum height, and $T$ is the time of flight, then:

$A$ 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 $g=10 \ ms^{-2}$].

One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to the horizontal. After two more seconds, it is travelling horizontally. Then the magnitude of the initial velocity of the projectile is $(g = 10 \,ms^{-2})$