Problem 2/95 Page 53, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Particle acted upon by gravity and a drag force proportional to the velocity.

Find: Velocity and position of the particle as a function of time.

Relevant variables:
m = Particle mass
V0 = Initial speed of particle
q = Angle of initial velocity with horizontal
K= m k = Drag force coefficient

1. Mechanical System: Particle traveling through fluid.

2. Free Body Diagram: To be completed.

3. Equations:
a.  S F = - m g j - K V
            = - m g j - K ( Vx i + Vy j )

b.  m a = m (Ax i + Ay j)
           = m ( dVx/dt i + dVy/dt j)

4. Solve:

m dVx/dt = -k Vx
m dVy/dt = -mg - K Vy

Dividing by m and recalling the definition of k,

dVx/dt = -k Vx
dVy/dt = -g - k Vy

Separating variables:

dVx / Vx = -k dt
dVy/ (g/k + Vy) = -k dt

Integrating both sides (note use of constants of integration) :

ln(Vx/Cx) = -k t
ln( (Vy + g/k)/Cy ) = -k t

Exponentiating both sides:

Vx = Cx e^-kt
Vy = -g/k + Cy e^-kt

Using the fact that at t=0, the initial values of the x and y components of the velocity are known to be V0 cos(q) and V0 sin(q):

Vx = V0 cos(q) e^-kt
Vy = ( V0 sin(q) + g/k ) e^-kt - g/k

Note that as t gets large, the exponential terms disappear and the terminal velocity becomes:

Vx -> 0      Vy -> -g/k

To determine the position as a function of time recall the definition of velocity:

V = d R / dt
Vx = dX/dt
Vy = dY/dt

Thus:

dX/dt = V0 cos(q) e^-kt
dY/dt = ( V0 sin(q) + g/k ) e^-kt - g/k

Separating variables:

dX = V0 cos(q) e^-kt dt
dY = ( V0 sin(q) + g/k ) e^-kt - g/k dt

Integrating both sides (note use of constants of integration)

X = -V0/k cos(q) e^-kt + Dx
Y = 1/k { ( V0 sin(q) + g/k ) e^-kt } - g/k t + Dy

Setting the origin of the coordinate system to be at the initial position of the particle so that X and Y must be zero when t=0, permits evaluation of the constants Dx and Dy.

X = V0/k cos(q) ( 1 -  e^-kt  )
Y = 1/k ( V0 sin(q) + g/k ) ( 1 - e^-kt  ) - g/k t

Isn't calculus wonderful?