The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

If a body is projected with an angle $\theta$ to the horizontal, then

If bullets are fired in all possible directions from the same point with an equal velocity of $10 \,m \,s^{-1}$ and an angle of projection $45^{\circ}$,then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,m^2$)

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)

$A$ body is projected from the ground at an angle of $\tan^{-1}(\sqrt{7})$ with the horizontal. At half of the maximum height, the speed of the body is '$n$' times the speed of projection. The value of '$n$' is

$A$ projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as the origin,the coordinates of the point where the speed of the particle is minimum are: