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# Vertical motion of a baseball under quadratic drag friction

Problem 1:

Consider a two-dimensional problem in the xy-plane. There are two points in space, labeled as P and Q, as shown in the figure to the right
a) Using polar coordinates, sketch on the figure at points P and Q the radial and azimuthal unit vectors.
b) Write down a general expression for the radial and azimuthal unit vectors in terms of the azimuthal angle and the Cartesian unit vectors i and j. Use the axis along the i direction as your reference line following standard practice.
c) Show by providing a complete derivation, how the derivatives with respect to the azimuthal angle of these two polar unit vectors relate to one another.

Problem 2:

In this problem we will analyze the motion of a baseball having mass of 0.15 Kg
and diameter of 7 cm. This baseball is thrown upwards with an initial speed of vo = 20 m/sec. On the surface of the Earth, g = 9.8 m/sec2 and the terminal speed of the baseball at STP conditions is: vter = 35m/sec which is ≈80 mi/hr. Read all parts of this problem carefully, and answer as many parts as you can before time runs out. Use y as the coordinate for the vertical motion, where + is taken upward. Since the ball's motion is along one-dimension (the y-axis) do not consider the x or z coordinates. Let vy denote the y component of
the velocity.

a) Neglect air friction altogether, and draw a free body diagram (FBD) with all forces labeled.
b) Assuming quadratic drag friction, draw the appropriate FBD when the ball is moving upward.
c) Assuming quadratic drag friction, draw the appropriate FBD when the ball is moving downward.
d) Write down Newton's differential equation for all three cases (a, b, c), but do NOT SOLVE them.
e) USING dvy/dt = vy dvy/dy ----- Determine vy(y) for case (a) when there is no drag friction. In other words, find vy as a function of position (y). You must take into account the given initial conditions.
f) From the equation you just derived in (e) --- describe (explain) its physical significance.
g) What is the formula for the maximum height the ball will travel upward from its point of release?
h) Based on the numbers given above, and the equation you just derived for part (g) --- What is the numerical value for the maximum height?
i) AGAIN USING dvy/dt = vy dvy/dy ----- Determine vy(y) for case (b) when there is quadratic drag friction as the ball is traveling upward. In other words, find vy as a function of position (y). You must take into account the given initial conditions. Hint, to solve this elementary integral (meaning looking it up is not necessary), consider a change of variable that looks something like u = 1 + x2. I will leave to your imagination on what "x" is, and how you will use this hint to solve this problem.
j) From answer (i) or using the same approach, write the formula for maximum height, ymax in terms of g, vo and vter.
k) From your answer in (j) show that in the limit of no friction, ymax reduces to the case you already solved (or should have solved) in part (g).
l) Based on the numbers given above, and the equation you derived for part (j) --- What is the numerical value for the maximum height? How does it compare to your answer in part (h).
m) USING dvy/dt = vy dvy/dy ----- Determine vy(y=0) for case (c) when the ball reverses its direction and free falls down with quadratic drag friction. In other words, find the final vy when the ball returns to its starting point. You must take into account the given (new) initial conditions. You do not need to calculate vy as a function of y in this problem or any other part of this test.
n) Find an approximate expression for the final vy assuming vyo/vter is a sufficiently small number to justify looking at the first nontrivial correction to the result you would obtain without air friction.

#### Solution Preview

Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here.

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(a) Diagram (see the attached file)

(b)

= i cos @ + j sin @
= - i sin @ + j cos @

(c)

d /d@ = - i sin @ + j cos @ =
d /d@ = - i cos @ - j sin @ = - (i cos @ + j sin @) = -

Unit vectors and are orthogonal. ( i.e. perpendicular to each other). Hence their dot product is zero.

. = 0

d /d@ . d /d@ = - . = 0

d /d@ . d /d@ = 0

This is the relationship between these two derivatives. Their scalar product (dot product) is zero.

2. Newton's second law says, F = m * acceleration

(a): Apply this law upward;

- mg = m a

-g = a = dv/dt

dv/dt = - g

(b): Apply this law upward;

- (mg+kv2) = m a

- (mg+kv2) = m dv/dt

dv/dt = - [g +(k/m) v2]

(c): Apply this law upward;

(mg - kv2) = m a

(mg - kv2) = m dv/dt

dv/dt = [g - (k/m) v2]

(e) dv/dt = - g

v dv/dy = -g

v dv = -g dy

integrating both sides,

v^2/2 = -g y + c

Boundary condition: at y = 0, v = vo

vo^2/2 = 0 + c

substituting the value ...

#### Solution Summary

Motion of an object thrown vertically up is discussed in detail taking in to account the drag friction. Every detail of the solution is provided for the student so that he/she will be able to answer similar questions in the future. Answer is in an 8-page word document.

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