The equation of motion of a projectile is $y = Ax - Bx^2$ where $A$ and $B$ are constants of motion. The horizontal range of the projectile is

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by

$A$ boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give an explanation to support your diagram.

$A$ stone is projected at an angle $\theta$ with velocity $u$. If it executes nearly a circular motion at its maximum point for a short time,the radius of the circular path will be ($g=$ acceleration due to gravity).

$A$ ball is thrown from a point on the ground at some angle of projection. At the same time,a bird starts from a point directly above this point of projection at a height $h$ and moves horizontally with speed $u$. Given that in its flight,the ball just touches the bird at one point. Find the distance on the ground where the ball strikes.

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.