Model Rocket Heights: Comparing a Quadratic Function and a Table
A physics class launches two model rockets from a rooftop. Rocket A's height in feet after t seconds is modeled by h_A(t) = -16t^2 + 64t + 80. Rocket B's height is recorded in a table: at t = 0, 1, 2, 3, 4, 5, and 6 seconds, the heights are 100, 120, 132, 136, 132, 120, and 100 feet. The class wants to know which rocket reaches the greater maximum height, which is higher at t = 4 seconds, and how the average rate of change from t = 0 to t = 2 compares for the two rockets. A student claims Rocket B must have a greater maximum because its table shows a height of 136 feet, while Rocket A's equation has a negative leading coefficient. The teacher asks the class to use function reasoning, not just the sign of a coefficient, to settle the comparison.
1. Find Rocket A's maximum height and the time it occurs. Then compare it with Rocket B's maximum from the table, and respond to the student's claim about the negative leading coefficient.
Read the worked answer
For Rocket A, (h_A(t)=-16t^2+64t+80). The maximum of a quadratic (at^2+bt+c) with (a<0) occurs at (t=-\frac{b}{2a}):
[t=-\frac{64}{2(-16)}=2.]
Substitute (t=2):
[h_A(2)=-16(4)+64(2)+80=-64+128+80=144.]
So Rocket A reaches 144 feet at 2 seconds. The table shows Rocket B reaches 136 feet at 3 seconds. Therefore Rocket A's maximum is greater by (144-136=8) feet.
The negative leading coefficient means Rocket A's graph opens downward, so it has a maximum rather than a minimum. It does not mean the maximum is lower. The vertex value determines the maximum, and Rocket A's vertex height is 144 feet.
2. Which rocket is higher at \(t=4\) seconds? Show calculations and explain what the table value means in context.
Read the worked answer
For Rocket A:
[h_A(4)=-16(4)^2+64(4)+80=-256+256+80=80\text{ feet}.]
The table gives Rocket B's height at (t=4) as 132 feet. The table entry ((4,132)) means that 4 seconds after launch, Rocket B is 132 feet above the ground. Since (132>80), Rocket B is higher at 4 seconds, by (132-80=52) feet.
This also shows Rocket A has already passed its maximum at (t=2) and is descending, while Rocket B is still near its maximum at (t=3).
3. Compute the average rate of change of each rocket's height from \(t=0\) to \(t=2\). Include units and explain which rocket rose faster on average over that interval.
Read the worked answer
For Rocket A, use (h_A(0)=80) and (h_A(2)=144). The average rate of change is
[\frac{144-80}{2-0}=\frac{64}{2}=32\text{ feet per second}.]
For Rocket B, the table gives (h_B(0)=100) and (h_B(2)=132). The average rate of change is
[\frac{132-100}{2-0}=\frac{32}{2}=16\text{ feet per second}.]
Rocket A rose faster on average over the first 2 seconds: 32 ft/s compared with 16 ft/s for Rocket B. The average rate of change is the slope of the secant line between the two points; it describes the overall change over the interval, not necessarily the speed at any single instant.