In the figure shown,the two projectiles are fired simultaneously. The minimum distance between them during their flight is ........ $m$.

$A$ body is moving with constant speed in a circle of radius $10 \ m$. The body completes one revolution in $4 \ s$. At the end of the $3^{rd}$ second,the displacement of the body (in $m$) from its starting point is:

$A$ particle moves along a parabolic path $y = 9x^2$ in such a way that the $x$-component of velocity remains constant and has a value of $\frac{1}{3} \; m/s$. The acceleration of the particle is $....... \; m/s^2$.

In the projectile motion of an object, the object reaches its maximum height where its speed is half of its initial speed. Then the ratio between the range and the maximum height of the projectile is

Derive the expressions for the time taken to reach maximum height,the total time of flight,and the maximum height attained by a projectile.

Show that for a projectile,the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$.
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$,where the symbols have their usual meanings.