$A$ ball thrown by one player reaches the other in $2 \, s$. The maximum height attained by the ball above the point of projection will be about .......... $m$.

$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

$A$ 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:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways, then the product $t_1 t_2$ is proportional to

If a particle of mass $m$ is projected at an angle $\alpha$ with the horizontal,then :