$A$ ball thrown by one player reaches another player in $2 \text{ s}$. The maximum height attained by the ball above the point of projection will be $.... \text{ m}$.

For a projectile motion,the angle between the velocity and acceleration is minimum and acute at

$A$ 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

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of:

$A$ ball projected from the ground at an angle of $45^o$ just clears a wall in front. If the point of projection is $4 \, m$ from the foot of the wall and the ball strikes the ground at a distance of $6 \, m$ on the other side of the wall,the height of the wall is ........ $m$.

$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by: