$A$ 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:

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The velocity with which the projectile is projected is ......... $m/s$.

What is the path followed by a moving body,on which a constant force acts in a direction other than the initial velocity (i.e.,excluding parallel and antiparallel directions)?

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal,is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$,it is away from the target by $15 \ m$. Then the angle of projection to hit the target is:

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............