At what angle should a particle be projected to cover the maximum range (in $^{\circ}$)?

If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be

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

$A$ projectile is fired at $30^{\circ}$ to the horizontal. The vertical component of its velocity is $80 \; ms^{-1}$. Its time of flight is $T$. What will be the velocity of the projectile at $t = \frac{T}{2}$?

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

$A$ boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle of $30^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\, m$ from the ground? $\left[ g = 10\, m/s^2, \sin 30^o = \frac{1}{2}, \cos 30^o = \frac{\sqrt{3}}{2} \right]$