Galileo writes that for angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal ranges described by the projectile are in the ratio of (if $\theta \le 45^\circ$):

$A$ 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 $\,m$)

$A$ 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. . . . .

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$,where $h$ is the height,$x$ is the horizontal distance,and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is: (projection speed = $u$,angle of projection from horizontal = $\theta$)

$A$ cricket ball is thrown by a player at a speed of $20\,m/s$ in a direction $30^{\circ}$ above the horizontal. The maximum height attained by the ball during its motion is $........\,m$ $\left( g = 10\,m/s^2 \right)$