The ratio between the maximum range and the square of the time of flight in projectile motion is:

Range of a bullet fired at $45^o$ to the horizontal is $980 \, m$. If the bullet is fired at the same angle from a car travelling horizontally at $18 \, km/h$ towards the target,then the range will be increased by:

Two projectiles are thrown with the same range $R$. If their maximum heights are $h_1$ and $h_2$ respectively,then what is the value of the range $R$?

$A$ body $P$ is projected at an angle of $30^{\circ}$ with the horizontal and another body $Q$ is projected at an angle of $30^{\circ}$ with the vertical. If the ratio of the horizontal ranges of the bodies $P$ and $Q$ is $1: 2$,then the ratio of the maximum heights reached by the bodies $P$ and $Q$ is

$A$ particle of mass $1 \ kg$ is projected with an initial velocity $10 \ ms^{-1}$ at an angle of projection $45^{\circ}$ with the horizontal. The average torque acting on the projectile,between the time at which it is projected and the time at which it strikes the ground,about the point of projection in $Nm$ is:

$A$ particle is projected from the ground with velocity $u$ at an angle $\theta$ with the horizontal. The horizontal range,maximum height,and time of flight are $R, H$,and $T$ respectively. They are given by $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$. Now,keeping $u$ fixed,$\theta$ is varied from $30^o$ to $60^o$. Then,