$A$ particle covers $50\, m$ distance when projected with an initial speed. On the same surface,it will cover a distance of ......... $m$ when projected with double the initial speed.

$A$ body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory,when it makes an angle $\left(\frac{\theta}{2}\right)$ with the horizontal is ($g$ = acceleration due to gravity).

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

$A$ projectile covers double the range compared to its maximum height attained. The angle of projection is

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to: