A shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection then the angular elevation of the summit $\beta$ is
$\frac{1}{2} \alpha$
$tan^{-1}(1/2)$
$tan^{-1}(1/2 \,\,tan \,\, \alpha )$
$tan^{-1}(2 \,\,tan\,\, \alpha )$
Two particles are moving along two long straight lines, in the same plane, with the same speed $= 20 \,\,cm/s$. The angle between the two lines is $60^o$, and their intersection point is $O$. At a certain moment, the two particles are located at distances $3\,m$ and $4\,m$ from $O$, and are moving towards $O$. Subsequently, the shortest distance between them will be
A projectile is fired with a velocity at right angle to the slope which is inclined at an angle $\theta$ with the horizontal. The expression for the range $R$ along the incline is
A projectile is projected with velocity of $25\, m / s$ at an angle $\theta$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\theta$ will be : [use $g =10 m / s ^{2}$ ]
The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is
The equation of a projectile is $y =\sqrt{3} x -\frac{ gx ^2}{2}$ the angle of projection is