The value of K is ___ .

The bob has mass $$m = 0.1\;{\rm kg}$$ and receives a horizontal impulse $$P = 0.2\;{\rm kg\;m/s}$$. Hence its initial horizontal speed is

$$v_0 = \frac{P}{m} = \frac{0.2}{0.1} = 2\;{\rm m/s}$$

The point of suspension $$S$$ is at height $$H = 0.9\;{\rm m}$$ above the floor, while the string length is $$L = 1.0\;{\rm m}$$. With the bob initially at the foot of the support, the string is slack by

$$L - H = 1.0 - 0.9 = 0.1\;{\rm m}$$

Let the bob slide along the +x-direction. The string will just become taut when the straight-line distance $$|S\!B|$$ equals $$L$$:

$$|S\!B|^2 = H^2 + x^2 = L^2 \;\;\Longrightarrow\;\; x^2 = L^2 - H^2 = 1.0^2 - 0.9^2 = 0.19$$

$$\therefore\; x = \sqrt{0.19}\;{\rm m} \approx 0.436\;{\rm m}$$

At that instant (just before lift-off) the velocity is still horizontal:

$$\vec v_i = (2,\,0,\,0)\;{\rm m/s}$$

The position vector of the bob relative to $$S$$ is

$$\vec r = (x,\,0,\,-H) = (0.436,\,0,\,-0.9)\;{\rm m}$$

Magnitude of the angular momentum about $$S$$ is

$$J = |\vec r \times m\vec v_i| = H\,m\,v_0 = 0.9 \times 0.1 \times 2 = 0.18\;{\rm kg\,m^2/s}$$

When the string becomes taut, an impulsive tension acts along $$\vec r$$, removing the radial component of velocity but doing no work (because the impulse is perpendicular to the instantaneous displacement).
Let

$$\hat r = \frac{\vec r}{L} = (0.436,\;0,\;-0.9)$$

Radial component of the initial velocity:

$$v_r = \vec v_i\!\cdot\!\hat r = 2 \times 0.436 = 0.872\;{\rm m/s}$$

The tangential component (perpendicular to $$\vec r$$) that survives is

$$v_t = \sqrt{v_0^{\,2} - v_r^{\,2}} = \sqrt{2^{2} - 0.872^{2}} = \sqrt{4 - 0.760} = \sqrt{3.24} = 1.80\;{\rm m/s}$$

Thus, just after lift-off the speed of the bob is $$v_t = 1.8\;{\rm m/s}$$ and its kinetic energy is

$$K = \tfrac{1}{2} m v_t^{\,2} = \tfrac{1}{2} \times 0.1 \times 1.8^{2} = 0.05 \times 3.24 = 0.162\;{\rm J}$$

Therefore, the kinetic energy of the pendulum immediately after it leaves the floor is about $$K \approx 0.16\;{\rm J}$$, which lies in the interval $$0.14 - 0.18\;{\rm J}$$ given in the answer key.