The equation of a projectile is given by $y = Px - Qx^2$,where $P$ and $Q$ are constants. The ratio of the maximum height to the range of the projectile is

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

$A$ 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:

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$