$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

$A$ bullet is fired at time $t=0$ with velocity $20 \ m/s$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time $0.1 \ s$ is (Assume $g=10 \ m/s^2$)

Two projectiles thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively,reach the maximum height in the same time. The ratio of their initial velocities is

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.

If the height of a projectile at a time of $2 \ s$ from the beginning of motion is $60 \ m$,then the time of flight of the projectile is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be