Refer to figure.
Learning objective 081.03.01.02.02: Solve VS1G from the lift formula given varying C_L.

 The 1g stall speed (V_S1G) is the airspeed at which, for a load factor of 1, we will reach the critical angle of attack. At this critical angle of attack, we will have the largest C_L possible - C_LMAX. From the lift equation (L = ½ ρ V² S C_L) we see that lift is proportional to V². For a C_L fixed at C_LMAX if we want to create more lift, we need to be flying faster.

With a load factor of 1, Lift = Weight = mg 
Where m is the mass of the aircraft, and g is the acceleration due to gravity (here said to be 10 m/s²)

We can read off C_LMAX from the question annex - it is the maximum C_L, which coincides with the critical angle of attack. In this case, C_LMAX = 1.5

We can now rewrite the lift formula in terms of V_S1G, C_LMAX and aircraft mass, and then re-arrange to solve for V_S1G:

• L = ½ ρ V² S C_L
• L = W = m g = ½ ρ V_S1G² S C_LMAX
• 2 V_S1G²  = m g / (ρ S C_LMAX)
• V_S1G = √ (2 m g / (ρ S C_LMAX))

Substituting values, we find:

• V_S1G = √ (2 x 2000 x 10) / (1.225 x 16 x 1.5)) = 36.9 m/s

Caution!: The answers are in knots, not m/s! We must convert kt to m/s
We can use a conversion factor of 1 kt = 0.514 m/s:

• 36.9/0.514 = 71.8 kt