does anyone know how to do this accurately? bouncing ball question?

Question

does anyone know how to do this problem??

I have done the sketches, but I'm confused because it says to "show all work" and I can't really think of any way to do it accurately... I just guessed the general shapes of the graphs.

Please help if you have any idea!!

1. Imagine dropping a ball onto a hard floor so that it bounces straight up several times. Consider the time interval that the ball spends in contact with the floor to be of short duration but now zero.
a. sketch a position vs time graph for the motion for > 3 bounces. Assume that after each bounce the ball returns to a smaller maximum height. Take up to be the positive direction and the level of the floor to be zero height.
b. Directly below the graph from part (a) and using the same scale for the time axis, sketch a velocity vs time graph for the motion of the ball.
c. Directly below the graph for part (b) and using the same scale for the time axis, sketch an acceleration vs time graph for the motion of the ball.

Answer 1

Start with the ideal equations of motion for the ball:

When falling, y(t)=h-.5*g*t^2 for 0<=t<=t1
The time it strikes the floor is t1, which we can calculate using the y(t) equation
It will strike the floor at a speed we can calculate using conservation of energy, where
.5*m*v^2=m*g*h
or
v=sqrt(2*g*h)

In an elastic collision with the floor it would bounce back up with the same return speed upward
y(t)=v*(t-t1)-.5*g*(t-t1)^2 for t1<=t<=t2
note that at t=t1, y(t)=0
The second root for t gives us the time it will strike the floor again, t2
Under ideal and elastic conditions, the ball will return to the same height, h, and strike the floor with speed v

So the motion for y(t) for t2<=t<=t3
y(t)=v*(t-t2)-.5*g*(t-t2)^2 for t2<=t<=t3
t3 is when the ball strikes the floor again.

This is called a "piecewise linear" graph, where there are intervals of the independent variable, time in this case,
that the function of the dependent variable y(t), is linear.
The discontinuities are non-linear. In this case the bounces are the non-linear parts.

To finish off the problem ,assum that 10% of the kinetic energy of the ball is lost in each bounce
so v2=sqrt(.9*2*g*h);
and h2=.9*h, v3=sqrt(.81*2*g*h)
and that the bounce lasts for 0.1 seconds.
now the piecewise linear equations for three bounces are

y(t)=h-.5*g*t^2 for 0<=t<=t1
y(t)=0 for t1<=t<=t2, where t2 = t1+0.1 (first bounce)
y(t)=v2*(t-t2)-.5*g*(t-t2)^2 for t2<=t<=t3
y(t)=0 for t3<=t<=t4, where t4 = t3+0.1 (second bounce)
y(t)=v*(t-t4)-.5*g*(t-t4)^2 for t4<=t<=t5
when y(t)=0 for t>t5, the third bounce occurs


Let's use a h=1 m, g=10m/s^2, and calculate the times of each bounce
y(t1)=0=1-5*t1^2
t1=sqrt(1/5) or 0.5 seconds
v2=sqrt(.9*2*10*1) or 4.2 m/s
t2 =.6
the flight time between the first and second bounce is found with
0=4.24*t-5*t^2
the positive root is sqrt(5*4.24), or 0.9
so t3=0.6+0.9
t3=1.5
t4=1.6
v3=sqrt(.81*2*10*1)
v3=4.0 m/s
the flight time between the second and third bounce is found with
0=4.0*t-5*t^2
or 0.90
so t5=2.5

Now we're ready to graph position versus time.


For velocity, the equations are

v(t)=-g*t for 0<=t<=t1
v(t)=0 for t1<=t<=t2, where t2 = t1+0.1 (first bounce)
v(t)=v2-g*(t-t2) for t2<=t<=t3
v(t)=0 for t3<=t<=t4, where t4 = t3+0.1 (second bounce)
v(t)=v-g*(t-t4) for t4<=t<=t5

Here is the link to the graphs

http://i142.photobucket.com/albums/r88/odu83/bounce.jpg

and the data

http://i142.photobucket.com/albums/r88/odu83/bouncedata.jpg

j

Answer 2

A bouncing ball is usually considered to have a constant "coefficient of restitution" (which you can google for more info) meaning that on each bounce it loses a fixed proportion of its velocity. So on a given cycle the velocity (V) amplitude should be a constant proportion N (<1) of the previous cycle, and the position (P) amplitude (bounce height) should be N^2 * its previous value. The acceleration (A) pulse, assuming a constant duration of contact, would have N * its previous value. The duration of each cycle is proportional to V (since almost all the cycle is constant acceleration reducing V) and thus decreases cycle-to-cycle to N * its previous value. Obviously P, V and A are synced to each other (the start of the bounce being a possible point of reference). Note that the initial drop is just the same as if it were the last half of a preceding cycle.

So you have decreasing amplitudes and periods.
Position vs. time cycle is a parabola based at y=0. Each successive parabola should look exactly like an upper section of the previous one since its initial V corresponds to previous cycle's V at a later time.
Velocity vs. time is a "sawtooth" wave. Cycle consists of nearly vertical slope ("short-duration but not zero") up from previous cycle's final value -V(n-1) to new value V(n) computed for this cycle, then negative-going constant shallow slope to -V(n).
Accel. vs. time is a short-duration positive rectangular pulse, followed by constant -9.8 m/s^2. The area under the positive pulse equals the average of the areas under the preceding and following constant parts, so it's very tall relative to 9.81. (How tall depends on how short you define the bounce to be.)
EDIT: odu, your velocity trace is inconsistent. It should show energy loss (|V(n)| < |V(n-1)|) over the bounces and not over the free-flight phases. Also, I consider it a bit iffy to put in a placeholder ("I'll be back" or whatever you had) in order to be answerer #1, especially when the placeholder contained not a hint of an answer.