$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

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

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 angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)

$A$ particle is projected at an angle $\theta$ with the horizontal from the ground. The slope $(m)$ of the trajectory of the particle varies with time $(t)$ as:

The ceiling of a long hall is $25\; m$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\; m/s$ can go without hitting the ceiling of the hall (in $; m$)?