A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have same K.E. as that of boy. The original speed of the man will be

Let us solve the problem step-by-step by setting up equations for the kinetic energy of the man and the boy.

Let:
$M_m$ be the mass of the man,
$v_m$ be the original speed of the man,
$M_b$ be the mass of the boy, and
$v_b$ be the speed of the boy.

From the problem description, we are given that the boy has half the mass of the man:
$M_b=\frac{M_m}{2}$

The kinetic energy of any object is given by the formula:
$\mathrm{K.E.}=\frac{1}{2}m{v}^2$

Initially, the running man has half the kinetic energy of the boy:
$\frac{1}{2}{M}_m{v}_m^2=\frac{1}{2}\left(\frac{1}{2}{M}_b{v}_b^2\right)$

Substitute $M_b=\frac{M_m}{2}$ into the equation:
$\frac{1}{2}{M}_m{v}_m^2=\frac{1}{4}\left(\frac{M_m}{2}\right)v_b^2$

Simplify by canceling $M_m$ from both sides:
$\frac{1}{2}{v}_m^2=\frac{1}{8}$
Multiplying both sides by 8 gives:
$4v_m^2=v_b^2$
Taking the square root of both sides, we get the relationship between the speeds:
$v_b=2v_m$ (Equation 1)

Next, the man speeds up by 1 m/s (new speed becomes $v_m+1$) so that he has the same kinetic energy as the boy:
$\frac{1}{2}{M}_m{\left(v_m+1\right)}^2=\frac{1}{2}$

Substitute $M_b=\frac{M_m}{2}$ and cancel $\frac{1}{2}{M}_m$ from both sides:
${\left(v_m+1\right)}^2=\frac{1}{2}$ (Equation 2)

Substitute $v_b=2v_m$ from Equation 1 into Equation 2:
${\left(v_m+1\right)}^2=\frac{1}{2}$
${\left(v_m+1\right)}^2=\frac{1}{2}$
${\left(v_m+1\right)}^2=2v_m^2$

Taking the square root of both sides gives:
$v_m+1=\sqrt{2}{v}_m$

Rearranging the terms to solve for $v_m$:
$1=\sqrt{2}{v}_m-v_m$
$1=v_m\left(\sqrt{2}-1\right)$
$v_m=\frac{1}{\sqrt{2}-1}\mathrm{m/s}$

Therefore, the original speed of the man is $\left(\frac{1}{\sqrt{2}-1}\right)\mathrm{m/s}$.