Question

a smoke jumper jumps from a plane that is 1900 ft above the ground. the function $h = -16t^2 + 1900$ gives the jumper’s height $h$ in feet.
a. how long is the jumper in free fall if the parachute opens at 1000 ft?
b. how long is the jumper in free fall if the parachute opens at 940 ft?
c. what is a reasonable domain and range for the function $h$?

a. the jumper is in free fall for 7.50 s if the parachute opens at 1000 ft
(round to two decimal places as needed.)
b. the jumper is in free fall for 7.75 s if the parachute opens at 940 ft
(round to two decimal places as needed.)
c. what is a reasonable domain for the function $h$?

$\bigcirc$ a. $-1900 \leq t \leq 1900$
$\bigcirc$ b. $0 \leq t \leq 1900$
$\bigcirc$ c. $0 \leq t \leq 10$
$\bigcirc$ d. $-10 \leq t \leq 10$
$\bigcirc$ e. all real numbers

Explanation:

Step1: Analyze the function and domain context

The function \( h = -16t^2 + 1900 \) represents the height of the smoke jumper over time \( t \). Time \( t \) cannot be negative in this real - world context (we can't have negative time for the free - fall), so \( t\geq0 \). Also, the height \( h\) cannot be negative (the jumper can't go below ground level), so we find the time when \( h = 0\):
Set \( h = 0\), then \( 0=-16t^2 + 1900\).

Step2: Solve for t when h = 0

First, rearrange the equation: \( 16t^2=1900\).
Then, \( t^2=\frac{1900}{16}=\frac{475}{4} = 118.75\).
Taking the square root of both sides (and since \( t\geq0\)), \( t=\sqrt{118.75}\approx10.9\). But we can also check the options.
We know that \( t\) starts at \( t = 0\) (when the jumper starts to fall) and goes up to the time when the jumper reaches the ground. From the options, option A has negative time which is not possible. Option B has an upper limit of 1900 which is too large. Option D has negative time. Option E (all real numbers) is incorrect because time can't be negative and height can't be negative indefinitely. Option C: \( 0\leq t\leq10\) is a reasonable domain as it is non - negative time and within the time range before the jumper reaches the ground (since when \( t = 10\), \( h=-16\times10^2 + 1900=-1600 + 1900 = 300\) ft, still above ground, and as \( t\) approaches 10.9, \( h\) approaches 0).

Answer:

C. \( 0\leq t\leq10\)