A quadratic height model can reach the same height on the way up and on the way down. Set the model equal to the requested height, then graph both sides in Desmos to find the times where they meet. If the question says first, choose the earlier time; the later crossing is real, but it answers a different question.
Hints
- Hint 1
A quadratic equation can have two solutions because a projectile may pass the same height while rising and falling. Put the target height where is, and keep as the unknown time.
- Hint 2
Graph the modeled height and a horizontal line at . Each intersection, where the graphs meet, gives a time when the projectile is feet high. Which crossing comes earlier?
- Hint 3
Desmos shows approximate times, while the choices use exact radicals, or square-root expressions. Type a possible exact choice on a new line and compare its decimal with the earlier crossing.
Step-by-step
Graph the height and the target
Step 1Turn the target height into an equation
The model gives the projectile's height at time . Reaching feet means the height equals , so write:
- Step 2
Find both times on the graph
Type , using Desmos's for the time , and then type . Click both intersections, where the graphs meet. Desmos shows about and . The first coordinate of each point is a time when the projectile reaches feet.
- Step 3
Match the first time to an exact choice
First means use the smaller time, about seconds, not the later return to feet. Type on a new Desmos line. It shows about , matching the earlier crossing. The projectile first reaches feet seconds after launch. Choice D.