$A$ body is projected from the ground at an angle $\theta_1$ with the horizontal with kinetic energy $K$. What is the potential energy of the body when the velocity makes an angle $\theta_2$ with the horizontal?

Fill in the blanks:
$(a)$ For vectors $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} + \hat{j} - 2\hat{k}$,the $y$-component of the vector $(\vec{A} - \vec{B})$ is .....
$(b)$ The formula for the total time of flight of a projectile projected vertically upwards with an initial velocity $u$ is ..........
$(c)$ The formula for the centripetal acceleration of a particle moving in a circular path of radius $R$ with a constant angular velocity $\omega$ is ......
$(d)$ The component of any vector is always .........

In projectile motion,the modulus of the rate of change of velocity is:

$A$ particle is moving in a circle with uniform speed '$v$'. In moving from a point to another diametrically opposite point:

$A$ cyclist starts from the centre $O$ of a circular park of radius $1 \text{ km}$,reaches the edge $P$ of the park. Then,the cyclist cycles along the circumference to point $Q$ and returns to the centre along $QO$ as shown in the figure. If the round trip takes $10 \text{ minutes}$,the net displacement and average speed of the cyclist (in metre and kilometre per hour respectively) are:

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: