Answer:
\frac{4}{3}\pi R^3 \rho_s g = \frac{4}{3}\pi R^3 \rho g + \frac{1}{2} C_d \rho \pi R^2 w_s^2
\frac{4}{3}\pi R^3 \rho_s g = deposition of non-cohesive sediments
\pi = approx 3.14
R = radius of the spherical object being deposited
\rho = density of spherical object
g = gravitational acceleration
C_d = drag coefficient
w_s = particle's settling velocit
