$A$ ball is projected vertically downward with an initial velocity from a height of $20 \ m$ onto a horizontal floor. During the impact,it loses $50\%$ of its energy and rebounds to the same height. The initial velocity of its projection is ............ $m/s$.

$A$ body of mass $m$ moving with velocity $v$ collides inelastically with another body of the same mass. After the collision,the first body moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. What will be the speed of the second body after the collision?

Two identical balls $A$ and $B$ of equal mass $m$ are lying on a smooth surface as shown in the figure. If ball $A$ hits ball $B$ at rest with a velocity $16 \,ms^{-1}$,then the coefficient of restitution $e$ between $A$ and $B$ so that $B$ just reaches the highest point of the smooth inclined plane of height $5 \,m$ is $\left(g=10 \,ms^{-2}\right)$

$A$ ball falls freely from rest from a height of $6.25 \ m$ onto a hard horizontal surface. If the ball reaches a height of $81 \ cm$ after the second bounce from the surface,the coefficient of restitution is

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ strikes the rod at one end with a velocity $v$ perpendicular to the rod. If the particle comes to rest after the collision,the velocity of the center of mass of the rod after the collision is:

$A$ small steel ball bounces on a steel plate held horizontally. On each bounce, the speed of the ball arriving at the plate is reduced by a factor $e$ (coefficient of restitution) in the rebound, so that $V_{\text{upward}} = e V_{\text{downward}}$. If the ball is initially dropped from a height of $0.4 \text{ m}$ above the plate and if $10 \text{ s}$ later the bouncing ceases, the value of $e$ is: