$A$ projectile is projected at $10 \ m/s$ by making an angle of $60^{\circ}$ to the horizontal. After some time,its velocity makes an angle of $30^{\circ}$ to the horizontal. Its speed at this instant is:

$A$ projectile object is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with velocity of $140 \ m/s$. Then, the time after which its velocity makes an angle $30^{\circ}$ with the horizontal is (Take, $g=10 \ m/s^2$)

$A$ particle covers $50\, m$ distance when projected with an initial speed. On the same surface,it will cover a distance of ......... $m$ when projected with double the initial speed.

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40 \ m \ s^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to