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

$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ particle projected from the ground moves at an angle of $45^{\circ}$ with the horizontal one second after projection,and its speed is minimum two seconds after the projection. What is the angle of projection of the particle? [Neglect the effect of air resistance]

$A$ small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $v_0$. The velocity is averaged for the first $\sqrt{2} \,s$ and the magnitude of the average velocity comes out to be the same as that of the initial velocity,i.e.,$|v_0|$. The magnitude $|v_0|$ will be (take $g=10 \,m/s^2$):

$A$ projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) \, m/s$,where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 \, m/s^2$,the equation of its trajectory is:

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 $g = 10\, \text{m/s}^2$)