- The data below describe the various heights of a ball, in centimeters, after being bounced over the course of 1.06 seconds. I retrieved the data from the examples options given by Professor Little. How can I find the exact height of the ball at .63 seconds?
Time Height Time Height
- (Seconds) (cm) (Seconds) (cm)
0.0
|
212
|
0.59
|
87
|
0.03
|
182
|
0.63
|
49
|
0.07
|
147
|
0.66
|
2
|
0.1
|
103
|
0.69
|
30
|
0.13
|
55
|
0.73
|
63
|
0.17
|
0
|
0.76
|
89
|
0.2
|
40
|
0.79
|
108
|
0.23
|
80
|
0.83
|
120
|
0.26
|
114
|
0.86
|
129
|
0.3
|
141
|
0.89
|
126
|
0.33
|
161
|
0.92
|
121
|
0.36
|
177
|
0.96
|
105
|
0.4
|
182
|
0.99
|
86
|
0.43
|
182
|
1.02
|
63
|
0.46
|
178
|
1.06
|
27
|
0.5
|
164
| ||
0.53
|
146
| ||
0.56
|
118
|
- ARC of Secant Lines:
a) (49-87)(.63-.59)=-950
b) (49-118)(.63-.56)=-985.7
c) (49-146)(.63-.53)=-970
Because at this point the ball is falling, the slope is negative. depending on the time interval chosen, the rate at which the ball falls changes.
Point P: X= .63, Point Q: X= .6
IRC: (.6-.63) (.03)=-1second
The Instantaneous rate of change between points P and Q is -1, meaning that between those two points exactly the rate of change is -1. -1 is also the limit, or minimum of the line. Within the context of a bouncing ball, the derivative of the tangent line PQ is -1 cm.
SInce the y values (height) in the table fluctuate as the ball bounces, I know that the IRC = -1 because between points Q and P, the ball is falling and has a negative slope. As the ball continues to fall, and time passes, the slope of the secant lines get closer and closer to that of the tangent line.
Cool example. I like how you used a lot of points in your model.
ReplyDeleteI like the graph, very clear, and shows all of the points in the table
ReplyDeleterachel,
ReplyDeletei like the bouncing ball example! your graphs look good (except for the one i couldn't see) and your table looks good. for your initial question, you already have the height at 0.63 seconds, so your question probably should have been "what is the velocity at exactly 0.63 seconds?"
your secant calculations are accurate, but don't forget to include the units, which here would be ft/s. something must have happened with your tangent line calculation, because your value is about 900 units away from the calculations that you got for your secant lines. other than those few things, good job. =]
professor little