$A$ projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with the vertical and the range is $R_2$. Then:

Two guns $A$ and $B$ can fire bullets at speeds $1 \ km/s$ and $2 \ km/s$ respectively. From a point on a horizontal ground,they are fired in all possible directions. The ratio of the maximum areas covered by the bullets fired by the two guns on the ground is

$A$ particle is projected at an angle of $30^{\circ}$ from the horizontal at a speed of $60 \; m/s$. The height traversed by the particle in the first second is $h_0$ and the height traversed in the last second before it reaches the maximum height is $h_1$. The ratio $h_0 : h_1$ is . . . . . . . [Take $g = 10 \; m/s^2$]

The top of a smooth inclined plane of length $20 \sqrt{2} \,m$ is connected to the edge of a well of diameter $40 \,m$ making an angle $45^{\circ}$ with the vertical as shown in the figure. $A$ body is projected along the inclined plane with a velocity '$u$'. If the body crosses the well without falling into it, then the minimum value of '$u$' is $(g=10 \,ms^{-2})$.

The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

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