$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.

If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)

$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is: