At any instant $t$, the vertical distance $Y$ and horizontal distance $X$ of a projectile are given by $2Y = 6t - gt^2$ and $X = 4t$. The initial velocity of the projectile is ($X$ and $Y$ are in $m$ and $t$ is in $s$). (in $\,m/s$)

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 $(g = 10 \,ms^{-2})$

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

$A$ ball $A$ is projected vertically upwards with a certain initial speed $u$. Another ball $B$ of the same mass is projected at an angle of $30^{\circ}$ with the vertical with the same initial speed $u$. At the highest point,the ratio of the potential energy of ball $A$ to that of ball $B$ will be: $(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2})$

$A$ ball thrown by a boy is caught by another after $2 \ s$ at the same level some distance away. If the angle of projection is $30^o$,the velocity of projection is ......... $m/s$.

$A$ body is projected horizontally from the top of a tall tower with a velocity of $30 \,ms^{-1}$. At time $t_1$, its horizontal and vertical components of the velocity are equal and at time $t_2$, its horizontal and vertical displacements are equal. Then $t_2-t_1$ is (take, $g=10 \,ms^{-2}$ ) (in $\,s$)