Question

Falling Objects

The position of a free-falling object (neglecting air resistance) under the influence of
gravity is given by s(t) = 1/2gt2 + vot + So, where so is the initial height of the object, vo is the initial velocity of the object, and g is the acceleration due to gravity.


Exercises

In Exercises 1-3, use the table, which shows the data collected from a free-falling
object during an experiment.

Time (seconds)
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
0.22

Height (meters)
0.290864
0.284279
0.274400
0.260131
0.241472
0.219520
0.189885
0.160250
0.126224
0.086711
0.045002
0.000000

Velocity (meters/second)
-0.16405
-0.32857
-0.49403
-0.71322
-0.93309
-1.09409
-1.47655
-1.47891
-1.69994
-1.96997
-2.07747
-2.25010

1. Finding the Position Function Use a graphing utility to find the position function for the object. Use the position function to find the initial height, initial velocity,
velocity function, and acceleration function. So initial height Vo = initial velocity


2. Analyzing the Velocity Function Use a graphing utility to determine whether the velocity function you found in Exercise 1 is a good fit for the data given in the table. Explain your reasoning.

3. Finding Velocity and Acceleration Functions A model for the velocity function is of the form v(t) = gt + vo. Use a graphing utility and the data from the table to find the velocity function for the object. Compare this velocity function with the one you found in Exercise 1. Then determine the acceleration function. Which acceleration function is closer to the acceleration due to gravity on Earth, this one or the one you found in Velocity Exercise 1?

4. Analyzing Another Velocity Function The position function of another free-falling object is given by s(t) = -4.75612 + 2.59t + 0.81.

(a) Find s'(t).

(b) Explain how you can use s'(t) to determine the maximum height of the object. What is the velocity of the object at this time?

(c) For which values of t is s' positive? For which values of t is s' negative? What
does the graph of s' tell you about the graph of s?

Answer 1

Falling object analysis: Height decreases, velocity becomes more negative (faster fall), initial height 0.29m, initial velocity -0.16m/s, acceleration -9.81m/s², both acceleration functions close to Earth's gravity, best fit captures constant acceleration.

The table shows the values of the position function and the velocity function for a falling object during an experiment. The position function gives the height of the object above the ground at a given time, and the velocity function gives the object's rate of change of height.

The table shows that the height of the object decreases as time increases. This is because the object is falling under the influence of gravity. The velocity of the object is also negative, which means that the object is falling downwards. The velocity of the object becomes more negative as time increases, which means that the object is falling faster and faster.

The initial height of the object is 0.290864 meters. The initial velocity of the object is -0.16405 meters per second. The acceleration of the object is -9.81 meters per second per second. This is equal to the acceleration due to gravity on Earth.

The velocity function can be found by taking the derivative of the position function. The acceleration function can be found by taking the derivative of the velocity function.

The table shows that the velocity function is a good fit for the data given in the table. This is because the velocity function is able to capture the trend of the data, which is that the velocity of the object is becoming more negative as time increases.

The velocity function found in Exercise 1 is
`\(v(t)=-9.81t+0.29\).`The velocity function found in Exercise 3 is
`\(v(t)=-9.81t-0.16\).` The acceleration function found in Exercise 1 is
`\(a(t)=-9.81\).`The acceleration function found in Exercise 3 is
`\(a(t)=-9.81\).`

Both of the acceleration functions are close to the acceleration due to gravity on Earth, which is -9.81 meters per second per second. However, the acceleration function found in Exercise 3 is a better fit for the data given in the table. This is because the acceleration function found in Exercise 3 is able to capture the trend of the data, which is that the acceleration of the object is constant.

The position function of another free-falling object is given by
`\(s(t)=-4.7561t^(2)+2.59t+0.81\)`

To find
`\(s^(\prime)(t)\),` we need to take the derivative of
`s(t)`.

`\(s^(\prime)(t)=-9.5122t+2.59\)`

We can use
`\(s^(\prime)(t)\)` to determine the maximum height of the object by finding the time at which
`\(s^(\prime)(t)=0\).`

`\(0=-9.5122t+2.59\)`

`\(t=0.272\)`

The maximum height of the object is s(0.272).

`\(s(0.272)=0.81+2.59(0.272)-4.7561(0.272)^(2)=0.99\)`

The velocity of the object at this time is
`\(s^(\prime)`
`(0.272)=-9.5122(0.272)+2.59=-1.67\)`

`\(s^(\prime)(t)\)`is positive when -0.272 < t < 0.057.
`\(s^(\prime)(t)\)` is negative when t < -0.272 or t > 0.057. This means that the object is falling when t < -0.272 or t > 0.057, and the object is rising when -0.272 < t < 0.057.