If the range of a gun which fires a shell with muzzle speed $V$ is $R$,then the angle of elevation of the gun is

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of:

According to the quantum theory,a photon of electromagnetic radiation of frequency $v$ has energy $E = h v$,where $h$ is known as Planck's constant. According to the theory of relativity,a particle of mass $m$ has equivalent energy $E = m c^2$,where $c$ is the speed of light. Thus,a photon can be treated as a particle having effective mass $m = \frac{h v}{c^2}$. If a flash of light is sent horizontally in the Earth's gravitational field,then photons while travelling a horizontal distance $d$ would fall through a vertical distance given by:

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

$A$ ball is thrown upwards at an angle of $60^o$ to the horizontal. It falls on the ground at a distance of $90 \,m$. If the ball is thrown with the same initial velocity at an angle $30^o$,it will fall on the ground at a distance of ........ $m$.

$A$ particle is projected from level ground. Its kinetic energy $K$ changes due to gravity such that $\frac{K_{\max}}{K_{\min}} = 9$. The ratio of the range $R$ to the maximum height $H$ attained during its flight is: