A rigid slender bar, AB is sliding against two mutually perpendicular frictionless walls as shown in figure. The velocity of point A in downward direction at a given instant is 6 m/s. At that instant, the magnitude of absolute velocity of midpoint G is ______ m/s (Rounded off to two decimal places)

The correct answer is 4.24.

From the given image, we can identify a rigid slender bar AB sliding along two mutually perpendicular frictionless walls. Point A is in contact with the vertical wall (y-axis) and point B is in contact with the horizontal floor (x-axis). A downward velocity vector labeled $v_A$ is shown at point A with a magnitude of 6 m/s. The angle between the bar and the horizontal floor is labeled as 45°, and point G represents the midpoint of the bar.

Let the position of point A be represented by the coordinates (0, y) and the position of point B be represented by (x, 0).
Since the length of the bar $L$ is constant, the relation between the coordinates is given by the Pythagorean theorem:
$x^2+y^2=L^2$

Differentiating this relation with respect to time $t$, we obtain:
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$
$x{v}_B+y(-v_A)=0$

Here, $v_B=\frac{dx}{dt}$ is the horizontal velocity of point B moving to the right, and $\frac{dy}{dt}=-v_A=-6\text{m/s}$ represents the downward velocity of point A.
Solving for $v_B$ gives:
$v_B=\frac{y}{x}{v}_A$

At the given instant, the angle of the bar with the horizontal floor is 45°. Therefore, we have:
$y=L\sin (45°)$
$x=L\cos (45°)$
Thus, the ratio $\frac{y}{x}=\tan (45°)=1$, which leads to:
$v_B=\left(1\right)\left(6\right)=6\text{m/s}$

The coordinates of the midpoint G of the bar are:
$x_G=\frac{x}{2}$
$y_G=\frac{y}{2}$

Differentiating the coordinates of G with respect to time gives the velocity components of the midpoint:
$v_{xG}=\frac{1}{2}\frac{dx}{dt}=\frac{v_B}{2}=\frac{6}{2}=3\text{m/s}$
$v_{yG}=\frac{1}{2}\frac{dy}{dt}=-\frac{v_A}{2}=-\frac{6}{2}=-3\text{m/s}$

The magnitude of the absolute velocity of the midpoint G is the vector sum of its components:
$v_G=\sqrt{{v_{xG}}^2+{v_{yG}}^2}$
$v_G=\sqrt{3^2+{(-3)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\text{m/s}$

Calculating the numerical value:
$v_G\approx 3\times 1.4142=4.2426\text{m/s}$

Rounding to two decimal places, we get:
$v_G=4.24\text{m/s}$