If air resistance is proportional to the square of the instantaneous velocity, then the velocity v of a mass m dropped from a given height is determined from m dv dt = mg − kv2, k > 0. Let v(0) = 0, k = 0.125, m = 4 slugs, and g = 32 ft/s2. (a) Use the RK4 method with h = 1 to approximate the velocity v(5). (Round your answer to three decimal places.) v(5) ≈ ft/s (b) Use a numerical solver to graph the solution of the IVP on the interval [0, 6]. The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (2.92, 21.9), and exits the window nearly horizontal at the approximate point (6, 22.6). The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (0.98, 22.4), and exits the window nearly horizontal at the approximate point (6, 22.6). The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (0.78, 31.9), and exits the window nearly horizontal at the approximate point (6, 32.0). The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (2.42, 31.5), and exits the window nearly horizontal at the approximate point (6, 32.0). (c) Use separation of variables to solve the IVP and then find the actual value v(5). (Round your answer to four decimal places.) v(5) = ? ft/s I want to know the answer of (c)

Question

If air resistance is proportional to the square of the instantaneous velocity, then the velocity v of a mass m dropped from a given height is determined from

m

dv

dt

= mg − kv², k > 0.

Let

v(0) = 0, k = 0.125, m = 4 slugs, and g = 32 ft/s².

(a)

Use the RK4 method with

h = 1

to approximate the velocity v(5). (Round your answer to three decimal places.)

v(5) ≈ ft/s

(b)

Use a numerical solver to graph the solution of the IVP on the interval [0, 6].

The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (2.92, 21.9), and exits the window nearly horizontal at the approximate point (6, 22.6).

The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (0.98, 22.4), and exits the window nearly horizontal at the approximate point (6, 22.6).

The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (0.78, 31.9), and exits the window nearly horizontal at the approximate point (6, 32.0).

The t v-coordinate plane is given. The curve begins at the origin, goes up and right becoming less steep, starts becoming nearly horizontal at the approximate point (2.42, 31.5), and exits the window nearly horizontal at the approximate point (6, 32.0).

(c)

Use separation of variables to solve the IVP and then find the actual value v(5). (Round your answer to four decimal places.)

v(5) = ? ft/s

I want to know the answer of (c)