$A$ ball is rolled off along the edge of a horizontal table with velocity $4 \, m/s$. It hits the ground after time $0.4 \, s$. Which of the following are correct?

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

Two particles,one at the centre of a circle of radius $R$,and another at a point $Q$ on the circle,start moving towards a point $P$ on the circle at the same time (see figure below). Both move with uniform velocities $\vec{V}_1$ and $\vec{V}_2$ respectively. They reach the point $P$ at the same time. If the angle between the velocities is $\theta$ and the angle subtended by $P$ and $Q$ at the centre is $\phi$ (as shown in the figure),then:

$A$ plane is revolving around the Earth with a speed of $100\, km/hr$ at a constant height from the surface of the Earth. The change in the velocity as it travels a half circle is ......... $km/hr$.

$A$ man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. $A$ central square of $50\, m \times 50\, m$ is filled with sand. Outside this square,he can walk at a speed of $1\, m/s$. In the central square,he can walk only at a speed of $v\, m/s$ $(v < 1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

If two bodies $A$ and $B$ are projected with the same velocity $u$ but with different angles $\theta_1$ and $\theta_2$ respectively with the horizontal such that both have the same range,then the ratio of the times of flight of the bodies $A$ and $B$ is: