The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$, respectively. Initial velocity of the projectile in $m/s$ is

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal,is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$,it is away from the target by $15 \ m$. Then the angle of projection to hit the target is:

$A$ projectile is thrown with an initial velocity $(a \hat{i} + b \hat{j}) \ m/s$,where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it,then

$A$ body projected at a certain angle $(\neq 90^{\circ})$ from the ground crosses a point in its path at a time $t_1 = 2.3 \ s$ and from there it reaches the ground after a further time $t_2 = 5.7 \ s$. The maximum height reached by the body is (Take $g = 10 \ ms^{-2}$) (in $m$)

$A$ ball is projected from the ground with a speed of $20\,m/s$ at an angle of $45^{\circ}$ with the horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$.

The time of flight of an object projected with speed $20 \, m/s$ at an angle $30^{\circ}$ with the horizontal is .... $s$.