$A$ particle is projected from the ground with a speed of $80 \,m/s$ at an angle of $30^{\circ}$ with the horizontal. The magnitude of the average velocity of the particle in the time interval $t=2 \,s$ to $t=6 \,s$ is ....... $m/s$. (Take $g=10 \,m/s^2$)

$A$ 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:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

In a spring gun having a spring constant $k = 100\, \text{N/m}$,a small ball $B$ of mass $m = 100\, \text{g}$ is placed in its barrel (as shown in the figure) by compressing the spring by $x = 0.05\, \text{m}$. $A$ box is placed at a distance $d$ on the ground so that the ball falls into it. If the ball leaves the gun horizontally at a height of $h = 2\, \text{m}$ above the ground,find the value of $d$ in meters. (Take $g = 10\, \text{m/s}^2$)

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

The velocity of projection of a body is increased by $2 \%$. Other factors remaining unchanged,what will be the percentage change in the maximum height attained? (in $\%$)