$A$ particle is moving along a curve. Then

$A$ boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored,the forces acting on the ball at the position $X$ are represented by

$A$ 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 $m$)

The position vector of a particle moving in a plane is given by $r = a \cos \omega t \hat{i} + b \sin \omega t \hat{j}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along the rectangular axes $X$ and $Y$; $a$, $b$, and $\omega$ are constants, and $t$ is time. The acceleration of the particle is directed along the vector:

$A$ body is moving with an initial velocity of $10 \sqrt{2} \ m/s$ in the north-east direction. If it is subjected to an acceleration of $2 \ m/s^2$ directed towards the south, then the velocity of the body after $5 \ s$ is:

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as a function of $\phi$ can be represented as: