$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,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

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

$A$ particle is projected at an angle $\theta$ with the horizontal from the ground. The slope $(m)$ of the trajectory of the particle varies with time $(t)$ as:

$A$ bomb is released from a horizontal flying aeroplane. The trajectory of the bomb is

Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at an angle of $60^\circ$ to the horizontal,the ratio of the maximum heights reached is

$A$ ball $P$ is projected at an angle of $60^{\circ}$ with the vertical with a certain initial speed $u$. Another ball $Q$ of the same mass $m$ as that of ball $P$ is projected vertically upwards with the same initial speed $u$. At the highest point, the ratio of the potential energy of ball $P$ to that of ball $Q$ is $(\sin 30^{\circ} = 0.5)$.