We saw that a differential equation describing the velocity v of a falling mass subject to air resistance proportional to the instantaneous velocity is

m*dv/dt = mg-kv,
where k > 0 is a constant of proportionality.

The positive direction is downward.
a. Solve the equation subject to the initial condition v(0) = v0.
b. Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass.
c. If the distance s, measured from the point where the mass was released above ground, is related to velocity v by ds/dt = v(t), find an explicit expression for s(t) if s(0) = 0.

a) v(t) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K

where K = k/m

v(t) = [((kv₀/m) - g) e⁻ᴷᵗ + g] (m/k)

b) Terminal velocity = mg/k

c) s(t) = (gt/K) - [(Kv₀ - g) e⁻ᴷᵗ]

where K = k/m

s(t) = (mgt/k) - [((kv₀/m) - g))e⁻ᴷᵗ]

Step-by-step explanation:

m(dv/dt) = mg-kv

Divide through by m

(dv/dt) = g - (k/m)v

Let k/m = K

dv/dt = g-Kv

dv/(g - Kv) = dt

∫ dv/(g - Kv) = ∫ dt

∫ dv/(Kv - g) = - ∫ dt

(1/K) In (Kv - g) = -t + c (where c = constant of integration)

In (Kv - g) = - Kt + Kc

At t = 0, v = v₀

In (Kv₀ - g) = Kc

c = (1/K) In (Kv₀ - g)

In (Kv - g) = - Kt + Kc becomes

In (Kv - g) = - Kt + In (Kv₀ - g)

In (Kv - g) = [In (Kv₀ - g)] - Kt

Kv - g = (Kv₀ - g) e⁻ᴷᵗ

Kv = (Kv₀ - g) e⁻ᴷᵗ + g

v(t) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K

K = k/m

v(t) = [((kv₀/m) - g) e⁻ᴷᵗ + g] (m/k)

b) At terminal velocity, dv/dt = 0

From the starting differential equation,

m(dv/dt) = mg-kv

mg - kv = 0

kv = mg

v = mg/k

c) v = ds/dt = [(Kv₀ - g) e⁻ᴷᵗ + g]/K

(ds/dt) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K

Kds = [(Kv₀ - g) e⁻ᴷᵗ + g] dt

∫ Kds = ∫ [(Kv₀ - g) e⁻ᴷᵗ + g] dt

Ks = -K((Kv₀ - g) e⁻ᴷᵗ) + gt

s(t) = (gt/K) - [(Kv₀ - g) e⁻ᴷᵗ]

K = k/m

s = (mgt/k) - [((kv₀/m) - g))e⁻ᴷᵗ]

angle q= 92 degree

step-by-step explanation:

the solution is in the picture

when a quadrilateral is inscribed in a circle, the opposite angles are supplement to each other. in other words, the sum of the 2 angle is 180 degree