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

When a body is thrown with a velocity $u$ making an angle $\theta $ with the horizontal plane, the maximum distance covered by it in horizontal direction is

A projectile is launched from the origin in the $xy$ plane ( $x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance. $r =\sqrt{ x ^2+ y ^2}$ from the origin is plotted against $x$, the resulting curves show different behaviours for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure below. For $\alpha_1, r ( x )$ keeps increasing with $x$ while for $\alpha_2$, $r(x)$ increases and reaches a maximum, then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at an angle $\alpha_c\left(\alpha_1 < \alpha_c < \alpha_2\right)$. The value of $\alpha_c$ is [ignore where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity]

A projectile is fired from the surface of the earth with a velocity of  $5 \,m s^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 \,m s^{-1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $\,m s^{-1}$) is 
(Given $g = 9.8 \,m s^{-2}$)