$A$ particle is projected with velocity $2 \sqrt{gh}$ at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2h$ from each other. The time taken by the particle to travel between these two walls is

Two balls $A$ and $B$ are projected at an angle of $45^{\circ}$ and $60^{\circ}$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball $A$ to that for ball $B$ is $(\sin 30^{\circ} = \cos 60^{\circ} = 1/2, \sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}, \sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2)$.

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range 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

The equation of a projectile is $y = ax - bx^2$. Its horizontal range is ......

$A$ ball is rolled off the edge of a horizontal table at a speed of $4\, m/s$. It hits the ground after $0.4\, s$. Which statement given below is true?