$A$ stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following:

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

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$,where $h$ is the height,$x$ is the horizontal distance,and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

$A$ stone is thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$. Then the time of flight of the stone will be

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)

$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)