The equations of motion of a projectile are given by $x = 36t \, m$ and $2y = 96t - 9.8t^2 \, m$. The angle of projection is

$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

$A$ projectile is thrown from a point in a horizontal plane such that the horizontal and vertical velocities are $9.8 \; m/s$ and $19.6 \; m/s$ respectively. It will strike the plane after covering a horizontal distance of ........ $m$.

For a projectile motion,the angle between the velocity and acceleration is minimum and acute at

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