$A$ projectile is thrown such that it achieves a maximum range $R$. What are the coordinates of the point where its velocity is minimum?

$A$ gun is aimed at a target in the line of its barrel. The target is released and allowed to fall under gravity at the same instant the gun is fired. The bullet will

$A$ body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \,m$ from the foot of a vertical pole of height $20 \,m$. The body just crosses the top of the pole and strikes the ground at a distance $s$ from the foot of the pole on the other side. Then, $s$ is: (in $\,m$)

If the initial velocity of a projectile is doubled,keeping the angle of projection the same,the maximum height reached by it will:

$A$ 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 $km$)

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$