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 projectile is thrown with velocity $U=20\ m/s ± 5\%$ at an angle $60^o.$ If the projectile falls back on the ground at the same level then ......... $m$ of following can not be a possible answer for range.
A projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with vertical and the range is $R_2$, then
A projectile is fired at an angle of $45^o $ with the horizontal . Elevation angle of the projectile at its highest point as seen from the point of projection, is
A body is projected with velocity $u$ making an angle $\alpha$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is
Two projectiles, one fired from surface of earth with velocity $10 \,m/s$ and other fired from the surface of another planet with initial speed $5\, m/s$ trace identical trajectories. The value of acceleration due to the gravity on the planet is ......... $m/s^2$