The trajectory of a projectile projected from the origin is given by the equation $y = x - \frac{2x^2}{5}$. The initial velocity of the projectile is:

$A$ ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected with the same speed at an angle of $30^{\circ}$ with the horizontal. At the highest point,the ratio of their potential energies 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 $(\sin 30^{\circ} = 0.5)$.

$A$ projectile object is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with velocity of $140 \ m/s$. Then, the time after which its velocity makes an angle $30^{\circ}$ with the horizontal is (Take, $g=10 \ m/s^2$)

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by:

$A$ body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \,m$ from the foot of a vertical pole of height $20 \,m$. The body just crosses the top of the pole and strikes the ground at a distance $s$ from the foot of the pole on the other side. Then, $s$ is: (in $\,m$)