$A$ stone is projected with a velocity $20 \sqrt{2} \, m/s$ at an angle of $45^{\circ}$ to the horizontal. The average velocity of the stone during its motion from the starting point to its maximum height is $.......... \, m/s$ (take $g = 10 \, m/s^2$).

$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is

Given that $u_x$ = horizontal component of initial velocity, $u_y$ = vertical component of initial velocity, $R$ = horizontal range, $T$ = time of flight, and $H$ = maximum height of a projectile. Match the following two columns:

Column $I$	Column $II$
$(A)$ $u_x$ is doubled, $u_y$ is halved	$(p)$ $H$ will remain unchanged
$(B)$ $u_y$ is doubled, $u_x$ is halved	$(q)$ $R$ will remain unchanged
$(C)$ $u_x$ and $u_y$ both are doubled	$(r)$ $R$ will become four times
$(D)$ Only $u_y$ is doubled	$(s)$ $H$ will become four times

If a particle of mass $m$ covers half of the horizontal circle with constant speed $v$,then the change in its kinetic energy is

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.

$A$ particle is moving such that its position coordinates $(x, y)$ are $(2 \, m, 3 \, m)$ at time $t = 0$,$(6 \, m, 7 \, m)$ at time $t = 2 \, s$,and $(13 \, m, 14 \, m)$ at $t = 5 \, s$. The average velocity vector $\vec{v}_{av}$ from $t = 0$ to $t = 5 \, s$ is: