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

$A$ particle has an initial velocity of $(3\hat i + 4\hat j) \; ms^{-1}$ and an acceleration of $(0.4\hat i + 0.3\hat j) \; ms^{-2}$. Its speed after $10 \; s$ is:

Fill in the blanks:
$(a)$ If $\overrightarrow A \cdot \overrightarrow B = AB$,then the angle between $\overrightarrow A$ and $\overrightarrow B$ is ............
$(b)$ The velocity of a projectile at its maximum height is ............ (Take the angle of projection as $\theta$).
$(c)$ The projection of $\widehat i - 2\widehat j + 4\widehat k$ on the $y$-axis is ............

$A$ particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it equals $4R$. The angle of projection,$\theta$,is then given by

$A$ ball of mass $160 \, g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 \, m/s$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $\left(g=10 \, m/s^{2}\right)$ (in $kg \cdot m^{2}/s$).

The second's hand of a watch has a length of $6\,cm$. The speed of the end point and the magnitude of the difference of velocities at two perpendicular positions will be: