The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it,$P$ is equal to

$A$ ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times $2 \ s$ and $8 \ s$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ particle is projected at an angle of $30^{\circ}$ from the horizontal at a speed of $60 \; m/s$. The height traversed by the particle in the first second is $h_0$ and the height traversed in the last second before it reaches the maximum height is $h_1$. The ratio $h_0 : h_1$ is . . . . . . . [Take $g = 10 \; m/s^2$]

The path of a projectile is given by the equation $y = ax - bx^2$,where $a$ and $b$ are constants,and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection,respectively. The maximum height attained by the projectile and the angle of projection are respectively:

An aeroplane is flying horizontally with a velocity of $600 \, km/h$ at a height of $1960 \, m$. When it is vertically above a point $A$ on the ground,a bomb is released from it. The bomb strikes the ground at point $B$. The distance $AB$ is

$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is