$A$ projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) \, m/s$,where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 \, m/s^2$,the equation of its trajectory is:

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 angle with the horizontal at which the projectile was projected is:

Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$, the maximum height reached by it is: (in meters)

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $= g$)

An object is projected from the ground with a speed of $20 \, m/s$ at an angle of $30^{\circ}$ with the horizontal. Its centripetal acceleration one second after the projection is .......... $m/s^2$ [Take $g = 10 \, m/s^2$].