Hints that were given are Newtons second law, bouncy force, drag force?

Question

A steel sphere in a tank of oil is given an initial velocity V= 2i (m/s) at the origin of the coordinate system shown (metal sphere is at the top of oil barrel). The radius of the sphere is 15mm The density of the steel is 8000 kg/m^3 and the density of the oil is 980 kg/m^3. If V is the sphere's volume, the (upward) bouncy force on the sphere is equal weight of a volume V of oil. The magnitude of the hydrodynamic drag force D on the sphere as it falls is |D| = 1.6|v| N, where |v| is the magnitude of the sphere's velocity in m/s.

A) What are the x and y components of the sphere's velocity at t = 0.1 s?

B) What are the x and y coordinates of the sphere at t = 0.1 s?

Any one who can answer either will receive points of even help me out !

Answer 1

Fx = - (1.6 Ns/m)Vx

Fy = ((4/3)πr^3)(ρs - ρo)g - (1.6 Ns/m)Vy

M = ((4/3)πr^3)ρs

Ax = dVx/dt = - (1.6 Ns/m)Vx / ((4/3)πr^3)ρs

Ay = dVy/dt = [((4/3)πr^3)(ρs - ρo)g - (1.6 Ns/m)Vy] / ((4/3)πr^3)ρs

dVx/Vx = - (1.6 Ns/m)dt / ((4/3)πr^3)ρs

dVx/Vx = - (1.2 Ns/m)dt / (π(8,000 kg/m^3)(0.015 m)^3)

dVx/Vx = - (400/9π Ns/m)dt

Vx = (2 m/s)e^-400t/9π

dX/dt = (2 m/s)e^-400t/9π

X = (9π/200 m)(1 - e^-400t/9π)

dVy/dt = [(1 - (980/8,000)(9.80665 m/s^2) - (400/9π Ns/m)Vy]

dVy/dt ≈ [8.605335375 - (400/9π Ns/m)Vy]

dVy/dt ≈ - 14.14711[Vy - 0.6082753]

dVy/[Vy - 0.6082753] ≈ - 14.14711dt

Vy - 0.6082753 ≈ [Vyo - 0.6082753]e^-14.14711t

Vy ≈ 0.6082753[1 - e^-14.14711t]

dY ≈ 0.6082753[1 - e^-14.14711t]dt

Y ≈ 0.6082753[t + (1 - (1/14.14711)e^-14.14711t]

A)

Vx(0.1) = (2 m/s)e^-40/9π

Vx(0.1) ≈ 0.48599 m/s

Vy(0.1) ≈ 0.6082753[1 - e^-1.414711]

Vy(0.1) ≈ 0.46047 m/s

B)

X(0.1) ≈ (9π/200 m)(1 - e^-40/9π)

X(0.1) ≈ 0.034353 m

Y(0.1) ≈ 0.6082753[0.1 + (1 - (1/14.14711)e^-1.414711]

Y(0.1) ≈ 0.65865 m