If the equation of motion of a projectile is $y=Ax-Bx^2$,then the ratio of the maximum height reached and the range of the projectile is

$A$ projectile is thrown at a speed which is twice its speed at its maximum height. If $R$ and $H$ are its range and maximum height respectively,then the ratio $\frac{R}{H}$ is

Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at an angle of $60^\circ$ to the horizontal,the ratio of the maximum heights reached is

$A$ 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 $m$)

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)