There are two points $P$ and $Q$ on a projectile with velocities $v_P$ and $v_Q$ respectively such that $v_P$ is perpendicular to $v_Q$ and $\alpha$ is the angle that $v_P$ makes with horizontal at point $P$. Find the correct option
$\tan \,\alpha \, = \,\frac{{{v_Q}}}{{{v_P}}}$
$\cos \,\alpha \, = \,\frac{{{v_Q}}}{{{v_P}}}$
$\sec \,\alpha \, = \,\frac{{{v_Q}}}{{{v_P}}}$
$\cot \,\alpha \, = \,\frac{{{v_Q}}}{{{v_P}}}$
A particle projected from ground moves at angle $45^{\circ}$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]
The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is
The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:
A body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up the velocity of the mass after $ t$ seconds will be
A projectile is thrown in the upward direction making an angle of $60^o $ with the horizontal direction with a velocity of $147\ ms^{-1}$ . Then the time after which its inclination with the horizontal is $45^o $ , is ......... $\sec$