Two bodies are thrown up at angles of $45^\circ$ and $60^\circ$,respectively,with the horizontal. If both bodies attain the same vertical height,then the ratio of the velocities with which these are thrown is:

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

The position vector of a particle is given as $\vec{r} = (t^2 - 8t + 12)\hat{i} + t^2\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ........ $sec$.

Show that for a projectile,the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$.
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$,where the symbols have their usual meanings.

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 ............

$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$, where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.