Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

7.6.2 The Force of Gravity Change in Quadratic Functions

7.6.2 • The Force of Gravity Change in Quadratic Functions

Activity

Earlier, in activity 7.6.1, you completed a table that represents the height of a cannonball, in feet, as a function of time, in seconds, if there was no gravity.

This table shows the actual heights of the ball at different times.

Seconds

0

1

2

3

4

5

Distance above ground (feet)

10

400

758

1084

1378

1640

1. Compare the values in this table with those in the table you completed earlier. What are two similarities or differences that you notice?

Compare your answer:

The actual height values are all less than the hypothetical ones (except the first, before the cannonball has been fired).
The difference between the distances after some number of seconds is not constant. For example, 1 second after firing, the distances in the two tables are 416 and 400, a difference of 16 feet. Two seconds after firing, the difference between the two values is 64 feet.

2. Use the graphing tool or technology outside the course.

Access multimedia content

a. Plot the two sets of data you have on the same coordinate plane.

Compare your answer:

b. How are the two graphs alike? How are they different?

Compare your answer:

The dots (triangles) that form a straight line represent the hypothetical data that ignore gravity. The dots that form a curve represent the actual heights. They both start at the same place (0,10). The hypothetical data show the heights increasing at a constant rate, in a linear pattern, while the actual data show values growing more slowly each successive second.

3. What is the starting position for both graphs?

The starting position for both graphs is 10.

4. What was the constant rate of change for the cannonball without gravity?

The constant rate of change for the cannonball without gravity was 406.

The constant rate of change is also the initial velocity. Recall that the effect of gravity is represented by $16 t^{2}$ . Gravity causes the bend in the curve for the second table and graph because it pulls the cannonball down toward the ground.

5. With a partner, write an equation to model the actual distance $d$ , in feet, of the ball $t$ seconds after it was fired from the cannon. Consider gravity and how the points changed on the graph.

Compare your answer:

Notice that gravity caused the heights to decrease, so, the effect of gravity is to subtract $16 t^{2}$ . An equation for the actual height of the cannonball is $d = 10 + 406 t - 16 t^{2}$ .

Self Check

An arrow is shot up into the air from 5 feet, at an initial velocity of 220 feet per second. Which equation can represent the height,

h(t)

, for any time,

t

?

$h(t)=220+5t-16t^2$
$h(t)=5+220t-16t^2$
$h(t)=5+220t+16t^2$
$h(t)=5+220t$

Additional Resources

The Effect of Gravity on Motion

A firework is sent up in the air, at an initial velocity of 130 feet per second, from a height of 6 feet. Write an equation that represents the height, $h (t)$ , of the firework for any second, $t$ . Remember to consider the effect of gravity.

Step 1 - Identify the initial velocity, $v_{0}$ .

130 ft per second

Step 2 - Identify the starting height, $h_{0}$ .

6

Step 3 - Write an equation of this form: $h (t) = h_{0} + v_{0} t - 16 t^{2}$ .

$h (t) = 6 + 130 t - 16 t^{2}$

Try it

The Effect of Gravity on Motion

Write an equation if a firework is shot from 2 feet off the ground at an initial velocity of 110 feet per second. Write an equation that represents the height, $h (t)$ , of the firework for any second, $t$ . Remember to consider the effect of gravity.

Here is how to write the equation for the height of the firework:

Step 1 - Identify the initial velocity, $v_{0}$ . 110 ft per second

Step 2 - Identify the starting height, $h_{0}$ . 2

Step 3 - Write an equation of this form: $h (t) = h_{0} + v_{0} t - 16 t^{2}$ . $h (t) = 2 + 110 t - 16 t^{2}$