Project 12. Approximate Height of a Dropped Object.
Using the acceleration due to gravity, 9.80665 m⁄s2, we can calculate the velocity at any time. However, we can only approximate its position, because the velocity is constantly changing. This activity demonstrates a method of approximation that is used in many problems including calculating areas (e.g., calculating the amount of asphalt needed for a driveway with curved sides). This activity also illustrates how approximations depend on choices we make.
(a)
Review the process shown here.
Suppose a ball is dropped from a height of 60.0 meters. We will approximate its height above ground by calculating its approximate position each second. For example, after 1 second the ball will have accelerated from 0 m/s to 9.8 m/s. If we pretend that it was travelling 9.8 m/s the whole time, then we estimate it will be 9.8 m lower than it began. This is a height of \(60.0-9.8=50.2\text{.}\) A second later the velocity will have increased 9.8 m/s so it will be \(9.8+9.8=19.6\) m/s. If we pretend that it was travelling 19.6 m/s the whole time, then we estimate it will be 19.6 m lower than it began. This is a height of \(50.2-19.6=30.6\text{.}\) Another second later (third second) the velocity will have increased by 9.8 m/s so it will be \(19.6+9.8=29.4\text{.}\) If we pretend that it was travelling 29.4 m/s the whole time, then we estimate it will be 29.4 m lower than it began. This is a height of \(30.6-29.4=1.2\text{.}\) With only 1.2 meters left and a velocity much greater than 1.2, we know it will reach the ground in less than one more second.
The table below contains these results.
| Time | Velocity | Distance Traveled | Height |
| 0 | 0 | 0 | 60.0 |
| 1 | 9.8 | 9.8 | 50.2 |
| 2 | 19.6 | 19.6 | 30.6 |
| 3 | 29.4 | 29.4 | 1.2 |
(b)
Repeat this exercise using half second intervals instead of one second intervals. Remember that in a half second the acceleration will be \(9.8/2 = 4.9\) m/s. In the first half second the ball will fall \(0.5 \text{ s} \cdot 4.9 \; \frac{\text{m}}{\text{s}} = 2.45 \text{ m}\text{.}\) In the second half second the velocity would be 9.8 m/s so it would drop \(0.5 \text{ s} \cdot 9.8 \; \frac{\text{m}}{\text{s}} = 2.45 \text{ m}\text{.}\) The first two entries are completed for you. Add more rows if you need to.
| Time | Velocity | Distance Traveled | Height |
| 0 | 0 | 0 | 60.0 |
| 0.5 | 4.9 | 2.45 | 57.55 |
| 1.0 | 9.8 | 4.9 | 52.65 |
| 1.5 | |||
| 2.0 | |||
| 2.5 | |||
| 3.0 |
(c)
The end result of this task will be a third table calculated in 1/4 second intervals.
(i)
What is the acceleration in 1/4 second? Recall it is 9.8 m/s which was 4.9 m/half second.
(ii)
Fill out a table in 1/4 second intervals.
(d)
Notice that in the first table (1 second intervals), we estimated after one second the height would be 50.2 m. In the second table (1/2 second intervals), we estimated after one second the height would be 52.65 m.
(i)
Find other times in which the 3 tables have different estimates.
(ii)
Are the estimates in one table always less than or greater than those in the other tables?
(iii)
Explain why these estimates are different.
(iv)
Which estimate do you think is most accurate?
