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mx" = 0
m/ = ~mg
Since x (0) = V0 cos Q0 and / (0) = V0 sin Q0, one integration of these second-order ODEs gives us
x(t) = V0 cos Q0 /(t) = V0 sin Q0 — gt Then because x(0) = 0 and /(0) = 6, a second integration yields
x(t) = (v0 cos Q0 )t
1 2 (2)
y(t) = 6 + Oosin<9o)f- 2^
To hit the target at some time T we want x(T) = 20 and /(T) = 10. So values of T > 0, Q0, and V0 such that
x( T) = 20 = (V0 cos Q0) T
y( D = 10 = 6 + Oosin0o) T- Ijgl2 ^
lead to hitting the target right in the bull’s eye and dunking your professor.
You can try to use system (3), or you can just adjust your launch angle and pitching speed by intuition and experience. The screen shot in Figure 5.1 shows you how to get started with the latter approach. If you play the dunking game on Screen 1.3 you’ll find that you can dunk without hitting the target head-on, but that a little up or a little down from the center works fine.
1The sin, cos and other trig functions in the ODE Architect Tool expect angles to be measured in radians. Note that 60 = 1 radian corresponds to 360/2^ ^ 57.3 degrees. The multimedia modules, however, will accept angles measured in degrees.
Longer to Rise or to Fall?
Figure 5.1: This ODE Architect screen shows paths of a ball thrown at ten different angles O0. Which paths lead to dunking the prof?
? Longer to Rise or to Fall?
Drag forces are usually determined by observation. They differ widely from one body to another.
Throw a ball straight up in the air and ask observers whether the ball takes longer to rise or to fall. You’ll get four answers:
1. Longer to rise
2. Longer to fall
3. Rise-time and fall-time are the same
4. It all depends ...
What’s your answer?
A mathematical model and ODE Architect suggest the answer. The forces acting on the ball of mass m are gravity and air resistance, so Newton’s second law states that
mR" (t) = —mgj + F
where R is the position vector of the ball, and F is the drag on the ball caused by air resistance. In this case R(t) = y(t)j where j is a unit vector pointing upward (the positive y direction). If the drag is negligible, we can set F = 0. For a light ball with an extended surface, like a whiffle ball, the drag, called viscous drag, exerts a force approximately proportional to the ball’s velocity but opposite in direction:
F(v) = -kv = —kyj
If the ball is solid and dense, like a baseball or a bowling ball, then we have Newtonian drag, which acts opposite to the velocity with magnitude proportional to the square of the speed:
F(v) = —k|v|v = —k|/1 yj
Summarizing, we have the models
/ = -g—
or, in system form,
y = v
ODE Architect only
accepts first-order ODEs, so vr = _____g____
that’s why we use the first-order system form.
To observe different rise times and fall times, you can set y(0) = 0, v(0) = V0 and see what happens for various positive values of V0. See Figure 5.2 for graphs of y( t) with viscous damping, four different initial velocities, k/ m = 2 sec—1, and g = 32 ft/sec2. In this setting v is the rate of change of y, so v is positive as the ball rises and negative as it falls.
0 no drag
k f ?
— y VISCOUS
— | y | y N ewtonian m
0 no drag
— v viscous m
— |u|u Newtonian m
? Indiana Newton
You notice that Indiana Newton is about to jump from a ledge onto a boxcar of a speeding train. His timing has to be perfect. He also gets to choose his drag: none, viscous, or Newtonian. If you knew the train’s position at all times, and how long it takes Indy to drop from the ledge to the top of the boxcar, then you could give him good advice about which drag to choose.
The initial value problem that models Indy’s situation is
y = v y(0) = h
v' = —g — F(v)/m v(0) = 0
where m is his mass, F(v) is a drag function, g is the acceleration due to gravity, and h is the height of the ledge above the boxcar. His life is in your hands! Figure 5.3 shows Indy’s free-fall solution curves y(t) from a height of 100 ft with three different drag functions.
Figure 5.2: Height vs. time of a whiffle ball thrown straight up four times with viscous damping and different initial velocities. Does the ball take longer to rise or to fall?
Figure 5.3: Indianajumps with no drag (left curve), viscous drag -0.2y (middle), Newtonian drag —0.02|/|y (right).
? Ski Jumping
Check that this force is perpendicular to velocity.
The origin of the wy-plane is at the edge of the ski jump (w-horizontal, y-vertical). The edge is horizontal so
W(0) = uq > 0, but ?(0) = 0.
When a ski jumper is aloft she is subject to gravitational, drag, and lift forces.
She can diminish the drag and increase the lift by her posture, ski angle, and
choice of clothing. Drag acts opposite to velocity and its magnitude is usually taken to be proportional to the skier’s velocity R':
Drag force = —SR' = —SXi — 5yj
The lift force is what makes ski jumping fun. The lift force is that force which acts perpendicular to the velocity and enables the jumper to soar. Its magnitude is usually taken to be proportional to the speed, so