Answer:
Equation of motion
x(t) = -9.81*t + 3000
The object will strike the ground in 305.81 sec
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