$A$ body is projected at $t=0$ with a velocity $10 \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1 \ s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10 \ m/s^2$,the value of $R$ is: (in $m$)

Assertion $(A)$: When a vehicle takes a turn on the road,it travels along a curved path. Reason $(R)$: In a curved path,the velocity of the vehicle remains the same.

$A$ particle moves in a plane along an elliptic path given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. At point $(0, b)$,the $x$-component of velocity is $u$. The $y$-component of acceleration at this point is

Two particles move on a circular path (one just inside and the other just outside) with angular velocities $\omega$ and $5\omega$ starting from the same point. Then

The position coordinates of a projectile projected from the ground on a certain planet (with no atmosphere) are given by $y = (4t - 2t^2) \text{ m}$ and $x = (3t) \text{ m}$,where $t$ is in seconds and the point of projection is taken as the origin. The angle of projection of the projectile with the vertical is ......... (in $^{\circ}$)

$A$ body of mass $2 \; kg$ has an initial velocity of $3 \; m/s$ along $OE$ and it is subjected to a force of $4 \; N$ in the $OF$ direction,which is perpendicular to $OE$. The displacement of the body from $O$ after $4 \; s$ will be: (in $; m$)