Two paper screens $A$ and $B$ are separated by a distance of $100\,m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $10\,cm$ below the hole in $A$. If the bullet is travelling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is $.......\,m/s$.

The top of a smooth inclined plane of length $20 \sqrt{2} \,m$ is connected to the edge of a well of diameter $40 \,m$ making an angle $45^{\circ}$ with the vertical as shown in the figure. $A$ body is projected along the inclined plane with a velocity '$u$'. If the body crosses the well without falling into it, then the minimum value of '$u$' is $(g=10 \,ms^{-2})$.

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)

$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at $45^\circ$ to the horizontal, then the ratio of maximum heights attained is