A quadratic height model can reach a height cutoff twice: once while rising and once while falling. Translate at least into an inequality, then graph the model with a horizontal cutoff line in Desmos. Find where the curve is on or above the line, and subtract the first crossing time from the second. The later crossing is an ending time, not a duration.
Hints
- Hint 1
The phrase at least includes heights equal to meters. Write an inequality comparing the rocket’s height with . At what height will the rocket enter and leave the qualifying range?
- Hint 2
A horizontal line at the cutoff height shows where the rocket first reaches that height and where it reaches it again on the way down. What do the time coordinates of those crossings mean?
- Hint 3
The two crossings mark the start and end of the time above the cutoff. A duration is found by subtracting the starting time from the ending time, not by using either time alone.
Step-by-step
Graph the height cutoff
Step 1Translate the height requirement
The words at least mean meters counts, so write the height requirement as an inequality:
- Step 2
Find the cutoff times
Type and in Desmos; stands for the problem’s time . Click the two intersections, where the graphs meet. Desmos shows and , so the rocket is at meters at 1 second and 3 seconds.
- Step 3
Identify when the height qualifies
The parabola, the curved graph of the quadratic model, is on or above the -meter line between those crossings. So the qualifying times are . Both endpoints count because the rocket is exactly meters high there.
- Step 4
Find the elapsed time
The question asks how long, not when the rocket last reaches meters. Elapsed time is the ending time minus the starting time:
The rocket is at least meters high for seconds. Choice B.