$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

$A$ 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).

If a body of mass $2 \,kg$ moving with an initial velocity of $4 \,m \,s^{-1}$ is subjected to a force of $3 \,N$ for a time of $2 \,s$ normal to the direction of its initial velocity, then the resultant velocity of the body is

$Assertion$ : When a particle moves in a circle with a uniform speed,its velocity and acceleration both change.
$Reason$ : The centripetal acceleration in circular motion is dependent on the angular velocity of the body.

$A$ sportsman runs around a circular track of radius $r$ such that he traverses the path $ABAB$. The distance travelled and displacement,respectively,are

$A$ particle is moving with velocity $\vec v = K(y\hat i + x\hat j)$ where $K$ is a constant. The general equation for its path is