Problem 2/49 Page 35, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

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

Find: Maximum height achieved by particle and the speed of the particle upon its return to launch point..

Relevant variables:
m = Particle mass
V0 = Initial speed of particle = 100 ft/s
K= m k = Drag force coefficient, k = 0.002 1/ft

First considering the upward travel of the particle:

1. Mechanical System: Particle traveling vertically upward.

2. Free Body Diagram: To be completed.

3. Equations:
a.  S F = - m g j - K V² j

b.  m a = m A j
           = m dV/dt j
           = m dV/dY dY/dt j
           = m dV/dY V j

4. Solve:

-m g - K V² = m V dV/dY

Dividing by m and recalling the definition of k:

-g - k V² =  V dV/dY

Separating variables:

-k dY = V dV / (g/k + V²)

Using the change of variable U = g/k + V²:

- k dY = 1/2 dU / U

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

-k Y = 1/2 ln (U/C)

Recalling the definition of U:

-k Y = 1/2 ln{(g/k + V²)/C}

Noting that at Y=0, V is V0, permits the determination of C:

-k Y = 1/2 ln{(g/k+V²)/(g/k+V0²)}

Noting that at the maximum height, h, the particle speed must be zero:

- k h = 1/2 ln { g/k / (g/k + V02) }

h = 1/(2k) ln ( 1 + V0² k /g )

Using the given values:

h = 250 ft ln ( 1 + 20/32.2) = 120.8 ft

Now consider the downward travel of the particle:

1. Mechanical System: Particle traveling vertically downward.

2. Free Body Diagram: To be completed.

3. Equations:
a.  S F = - m g j + K V² j

b.  m a = m A j
           = m dV/dt j
           = m dV/dY dY/dt j
           = m dV/dY V j

4. Solve:

-m g + K V² = m V dV/dY

Dividing by m and recalling the definition of k:

-g + k V² =  V dV/dY

Separating variables:

k dY = V dV / (-g/k + V²)

Using the change of variable U = -g/k + V²:

 k dY = 1/2 dU / U

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

k Y = 1/2 ln (U/D)

Recalling the definition of U:

k Y = 1/2 ln{(-g/k + V²)/D}

Exponentiating both sides:

V² - g/k = D e^2kY

Noting that at Y=h, V is 0, permits the determination of D:

V² - g/k = -g/k e^2k(Y-h)

At Y=0, the return speed, Vf can be determined from:

Vf² - g/k = -g/k e^-2kh

Vf = { g/k (1-e^-2kh) }^1/2

Using the given parameter values:

Vf = 16100^1/2 ft/s {1 - 1/(1+20/32.2) }^1/2 = 78.5 ft/s