Question

An object of mass 2020 kg is released from rest 30003000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant bequals=5050 ​N-sec/m, determine the equation of motion of the object. When will the object strike the​ ground? Assume that the acceleration due to gravity is 9.81 m divided by sec squared9.81 m/sec2 and let​ x(t) represent the distance the object has fallen in t seconds

Answer 1

Answer:

Equation of motion

x(t) = -9.81*t  + 3000

The object will strike the​ ground in 305.81 sec

Step-by-step explanation:

a) Determine the equation of motion of the object

If x(t) repsent the distance the object has fallen in t seconds, then we have x'(t) and x''(t) represent the speed and acceleration respectively.

Since the force due to air resistance is proportional to the velocity of the object with proportionality constant b = 50 N-sec/m, we have

F(t) = 50x'(t)  (1)

Due to Newton's 2nd law of motion (assuming the mass remains constant )

F(t) = mx''(t) = 50x''(t)

As the mass was released from rest, the acceleration is the gravity, and

F(t) = 50*(-9.81) = -490.5 N/m

The negative sign is due to the fact that the mass is falling downwards.

Replacing in (1) we get

-490.5 = 50x'(t) ===> x'(t) = -9.81 m/sec

and

x'(t) = -9.81 m/sec

This is a simple differential equation which can be solved directly by integration

x(t) = -9.81*t  + C  

where C is a constant, but we know that x(0) = 3000 m, so

C = 3000

We then have our final equation of motion which is

x(t) = -9.81*t  + 3000

b) When will the object strike the​ ground?

The object will strike the ground for the instant t for which x(t) = 0

x(t) = 0 ===> 9.81*t = 3000 ===> t = 3000/9.81 ===> t = 305.81 sec