Question 880·Medium·Math

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$h(t) = -5t^2 + 40t + 3$

The function $h$ models the height, in meters, of a firework above the ground $t$ seconds after it is launched.

Approximately how many seconds after launch does the model predict the firework will return to the ground?

For a quadratic height model, an object reaches the ground when its height is $0$ . Set the model equal to $0$ , then use a positive-time regression in Desmos to find the landing time. Round only after finding that time. The parabola also has a highest point, but the peak is not the landing, and a negative time would be before launch.

Hints

Hint 1
A zero of a height model is an input that makes its output $0$ . Returning to the ground means the height is $0$ , not that the firework has reached its highest point. What equation captures that?
Hint 2
A quadratic can have two zeros, but the question asks for seconds after launch. Restrict the Desmos regression to positive times so a solution from before launch cannot be mistaken for the landing.

Step-by-step

Find the positive zero
Step 1
Turn the landing into an equation
At the ground, the height is $0$ meters, so set the model’s output to $0$ :
$-5t^2+40t+3=0.$
A zero is a time when the model’s height is $0$ .
Step 2
Find the time after launch
Type $-5x_1^2+40x_1+3\sim0\{x_1>0\}$ in Desmos. The $\sim$ asks Desmos to solve the equation, and $x_1$ stands for time; the restriction keeps only times after launch. Under PARAMETERS, Desmos shows $x_1\approx8.0743$ seconds.
Step 3
Round the landing time
Round the positive time to the nearest hundredth: $8.0743\approx8.07$ . So the model predicts the firework returns to the ground about $8.07$ seconds after launch. Choice C.

Question 880·Medium·Math

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