Answer:
the mass of the brick is approximately m ≈ 173 gr
Explanation:
If the brick if thrown from a cliff , is likely that it had reached its terminal velocity by the moment it is near the ground. Then by Newton's first law
m*g - 1/2*ρ*C*v²*A = 0
m = 1/2*ρ*C*v²*A/g
where
ρ = density of air . We assume a density of 1.225 kg/m³
v= terminal velocity = 20 m/s
C= drag coefficient = 0.82 for a brick shape
g = gravity = 9.8 m/s²
A = area perpendicular to the ground. We assume that it the smaller area, since is the more stable alignment. Then for a standard brick of 225 mm x 112.5 mm x 75 mm → A = 112.5 mm * 75 mm = 8.43*10⁻³ m²
replacing values
m = 1/2*ρ*C*v²*A/g = 1/2*1.225 kg/m³*0.82*(20 m/s)²*8.43*10⁻³ m²/9.8 m/s² = 0.1728 Kg ≈ 173 gr
m ≈ 173 gr