$A$ particle is projected from a horizontal plane ($x-z$ plane) such that its velocity vector at time $t$ is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by:

$A$ helicopter flying horizontally with a velocity of $288 \ km/h$ drops a bomb. If the line joining the point of dropping the bomb and the point where the bomb hits the ground makes an angle $45^{\circ}$ with the horizontal, then the height at which the bomb was dropped is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

Two balls $A$ and $B$ are projected at an angle of $45^{\circ}$ and $60^{\circ}$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball $A$ to that for ball $B$ is $(\sin 30^{\circ} = \cos 60^{\circ} = 1/2, \sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}, \sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2)$.

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 $\,m/s$)

$A$ ball at point '$O$' is at a horizontal distance of $7 \ m$ from a wall. On the wall,a target is set at point '$C$'. If the ball is thrown from '$O$' at an angle $37^{\circ}$ with the horizontal,aiming at the target '$C$',but it hits the wall at point '$D$',which is at a vertical distance '$y_0$' below '$C$'. If the initial velocity of the ball is $15 \ m/s$,find $y_0$ (given $\cos 37^{\circ} = \frac{4}{5}$). (in $m$)

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)