Step-by-step Derivations:

1. Understanding the Equilibrium Condition

   A hot air balloon floats in equilibrium when the upward buoyant force ($F_b$) exerted by the surrounding air is equal to the total downward gravitational force (weight, $W$) of the balloon and its contents. The buoyant force is given by Archimedes' principle: $F_b = \rho_{\text{air}} V g$, where $\rho_{\text{air}}$ is the density of the surrounding air, $V$ is the volume of the balloon, and $g$ is the acceleration due to gravity. The weight of the balloon is $W = M g$, where $M$ is the total mass.

   At equilibrium, $F_b = W$, which implies: $\rho_{\text{air}} V g = M g$ $\rho_{\text{air}} V = M$

2. Applying the Condition to the Initial State

   Initially, the balloon is at an equilibrium height $h_1 = 100$ m with a total mass $M_1 = 480$ kg. The density of air at height $h$ is given by the formula $\rho(h) = \rho_0 e^{-h/h_0}$. So, the air density at $h_1$ is $\rho(h_1) = \rho_0 e^{-h_1/h_0}$.

   The equilibrium equation for the initial state is: $\rho(h_1) V = M_1$ $\rho_0 e^{-h_1/h_0} V = 480 \quad \cdots(1)$

3. Applying the Condition to the Final State

   After throwing out $N$ sandbags, each of mass 1 kg, the new mass of the balloon is $M_2 = 480 - N$. The balloon rises to a new equilibrium height $h_2 = 150$ m. The volume $V$ remains unchanged. The air density at this new height is $\rho(h_2) = \rho_0 e^{-h_2/h_0}$.

   The equilibrium equation for the final state is: $\rho(h_2) V = M_2$ $\rho_0 e^{-h_2/h_0} V = 480 - N \quad \cdots(2)$

4. Solving for N

   To find $N$, we can divide equation (2) by equation (1). This eliminates the unknown volume $V$ and the constant $\rho_0$. $\frac{\rho_0 e^{-h_2/h_0} V}{\rho_0 e^{-h_1/h_0} V} = \frac{480 - N}{480}$ $e^{-h_2/h_0} e^{h_1/h_0} = \frac{480 - N}{480}$ $e^{(h_1 - h_2)/h_0} = \frac{480 - N}{480}$

5. Substituting Numerical Values

   We are given: $h_1 = 100$ m $h_2 = 150$ m $h_0 = 6000$ m

   Substitute these values into the equation: $e^{(100 - 150)/6000} = \frac{480 - N}{480}$ $e^{-50/6000} = \frac{480 - N}{480}$ $e^{-1/120} = \frac{480 - N}{480}$

6. Using Approximation for the Exponential Term

   The exponent $x = -1/120$ is very small ($|x| \ll 1$). We can use the Taylor series approximation $e^x \approx 1 + x$ for small $x$. $1 - \frac{1}{120} \approx \frac{480 - N}{480}$

7. Final Calculation

   The equation can be rewritten as: $1 - \frac{1}{120} = 1 - \frac{N}{480}$ $\frac{1}{120} = \frac{N}{480}$ $N = \frac{480}{120}$ $N = 4$

Therefore, the number of sandbags thrown out is 4.