$A$ projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\,m/s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly $........\,m/s$.

The trajectory of a particle in projectile motion is given by $y = x - \frac{x^2}{80}$. Here, $x$ and $y$ are in meters. For this projectile motion, match the following with $g = 10 \, m/s^2$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

Two balls are projected from the same point simultaneously. The first ball is projected vertically upwards with velocity $u_1$ and the second ball at an angle of projection $60^{\circ}$ to the ground level with velocity $u_2$. Both the balls reach the ground simultaneously. The ratio of their velocities $u_1 : u_2$ is

The position vector of a particle is given as $\vec{r} = (t^2 - 4t + 6)\hat{i} + (t^2)\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ....... $sec$.

$A$ 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

As shown in the figure,there is a spring-block system. $A$ block of mass $500\,g$ is pressed against a horizontal spring fixed at one end to compress the spring by $5.0\,cm$. The spring constant is $500\,N/m$. When released,calculate the horizontal distance from the edge of the table where it will hit the ground $4\,m$ below the spring. $(g = 10\,m/s^2)$