$A$ particle moves in the $xy$-plane with velocity $v_x = 8t - 2$ and $v_y = 2$. If it passes through the point $(14, 4)$ at $t = 2 \, s$,the equation of its path is:

$A$ projectile of mass $200 \ g$ is launched in a viscous medium at an angle $60^{\circ}$ with the horizontal,with an initial velocity of $270 \ m/s$. It experiences a viscous drag force $\vec{F} = -c \vec{v}$,where the drag coefficient $c = 0.1 \ kg/s$ and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after $2 \ s$. Taking $e = 2.7$,the horizontal distance of the wall from the point of projection (in $m$) is

$A$ projectile is launched from the origin in the $xy$ plane ($x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance $r = \sqrt{x^2 + y^2}$ from the origin is plotted against $x$,the resulting curves show different behaviors for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure. For $\alpha_1$,$r(x)$ keeps increasing with $x$,while for $\alpha_2$,$r(x)$ increases and reaches a maximum,then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at a critical angle $\alpha_c$ (where $\alpha_1 < \alpha_c < \alpha_2$). The value of $\alpha_c$ is (where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity).

$A$ particle at a distance of $1 \ m$ from the origin starts moving,such that $dr/d\theta = r$,where $r$ and $\theta$ are polar coordinates. Then,the angle between the resultant velocity and the tangential velocity is

$A$ bomb is dropped from an aeroplane moving horizontally at a constant speed. When air resistance is taken into consideration,the bomb

The hour hand of a clock is $6\,cm$ long. The magnitude of the displacement of the tip of the hour hand between $1:00\,PM$ and $5:00\,PM$ is: