Question

A 200​-lb object is released from rest 600 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is minus10​v, where v is the velocity of the object in​ ft/sec, determine the equation of motion of the object. When will the object hit the​ ground? Assume that the acceleration due to gravity is 32 ft divided by sec squared and let​ x(t) represent the distance the object has fallen in t seconds.

Answer 1

Answer:

x(t) = 20t + 12.75e⁻¹•⁶ᵗ + 487.5

t = 24.375 s

Explanation:

The force balance on the object is given as

Net force = W - Drag force

ma = W - 10v

a = (dv/dt)

ma = m(dv/dt) = 200 - 10v

W = mg

200 = m×32

m = 6.25 kg

m(dv/dt) = 200 - 10v

6.25(dv/dt) = 200 - 10v

(dv/dt) = 32 - 1.6v

v' + 1.6v = 32

Solving this differential equation using the integrating factor method

(ve¹•⁶ᵗ) = ∫ (32e¹•⁶ᵗ) dt

ve¹•⁶ᵗ = (20e¹•⁶ᵗ) + c (where c = constant of integration)

v = (20 + ce⁻¹•⁶ᵗ)

At t = 0, v = 0

0 = 20 + c

c = -20

v = (20 - 20e⁻¹•⁶ᵗ)

v = (dx/dt)

(dx/dt) = 20 - 20e⁻¹•⁶ᵗ

dx = (20 - 20e⁻¹•⁶ᵗ) dt

x(t) = 20t + 12.5e⁻¹•⁶ᵗ + c (c is still the constant of integration)

At t = 0, x = - 500

- 500 = 0 + 12.5 + c

c = 512.5

x(t) = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

when the object hits the ground, x = 0

0 = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

20t + 12.75e⁻¹•⁶ᵗ = 487.5

Solving by trial and error,

t = 24.375 s

Hope this Helps!!!