For a projectile,the maximum height and horizontal range are same. The angle of projection $\theta$ of the projectile is

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

$A$ 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 $m$)

The angular momentum of a projectile projected at an angle $\theta$ with the horizontal with speed $u$ about the point of projection when it is at the highest point of its trajectory is

$A$ person can throw a stone to a maximum height of $h$. Determine the maximum horizontal range of the stone in terms of $h$.

$A$ projectile has a range of $1.5 \ km$ when projected at an angle of $15^{\circ}$. Find the range of the projectile when it is projected at an angle of $45^{\circ}$ with the same velocity. (in $km$)