Question

a leaf hangs from a branch 12 feet in the air. it falls to the ground at a rate of 0.25 feet per second. which graph could represent the leaf’s height in feet as a function of time, in seconds, after leaving the branch? four graphs labeled falling leaf with axes: height (feet) vs. time (seconds) are shown, with different lines and scales. the first has a steep line starting above 20, the second starts at 12, the third starts at 16, the fourth starts at 12.

Explanation:

Step1: Determine initial height

The leaf starts at 12 feet, so the graph should have a y - intercept of 12.

Step2: Determine rate of change

The leaf falls at 0.25 feet per second, so the slope is - 0.25. The time to reach the ground is $t=\frac{12}{0.25}=48$ seconds? Wait, no, wait. Wait, no, the formula for height $h(t)=12 - 0.25t$. When $h(t) = 0$, $0=12-0.25t$, so $t = 48$? But the graphs have time up to 20. Wait, maybe I misread. Wait, no, maybe the options are mis - presented? Wait, no, looking at the graphs:

First graph: starts above 12, wrong.

Second graph: starts at 12, slope is $\frac{2 - 12}{20-0}=\frac{- 10}{20}=-0.5$, no.

Third graph: starts at 16, wrong.

Wait, maybe the third graph is mis - labeled? Wait, no, the second graph (middle top) starts at 12. Wait, let's recalculate the time. Wait, $h(t)=12 - 0.25t$. Let's find when $h(t)=0$: $t = \frac{12}{0.25}=48$ seconds. But the graphs have time up to 20. Wait, maybe the problem has a typo, or I misread the rate. Wait, maybe the rate is 1 foot per second? No, the problem says 0.25. Wait, maybe the graphs are different. Wait, the second graph (middle top) has a y - intercept of 12, and the slope is $\frac{2 - 12}{20}=\frac{-10}{20}=-0.5$. The fourth graph (bottom middle) also starts at 12. Wait, maybe the correct graph is the one that starts at 12 (y - intercept 12) and has a slope of - 0.25. Let's check the second graph (middle top): when t = 0, h = 12; when t = 20, h=12-0.2520 = 12 - 5 = 7? No, the second graph at t = 20 has h = 2. Wait, no. Wait, maybe the rate is 1 foot per second? Then t = 12 seconds. Ah! Maybe I misread the rate. If the rate is 1 foot per second, then $h(t)=12 - t$, and when h = 0, t = 12. Let's check the third graph (right top): it starts at 16, no. Wait, the first graph: starts above 12, wrong. The second graph (middle top) starts at 12, and at t = 12, h=12-0.2512 = 12 - 3 = 9? No. Wait, maybe the correct graph is the one that starts at 12 (y - intercept 12) and has a slope of - 1 (if rate is 1). Then at t = 12, h = 0. The third graph (right top) starts at 16, wrong. Wait, maybe the problem's rate is 1 foot per second. Then the correct graph is the one where at t = 12, h = 0, and starts at 12. Wait, the second graph (middle top) starts at 12, but at t = 12, h is not 0. The third graph (right top) starts at 16, wrong. Wait, maybe the first graph is wrong, the second graph (middle top) is the only one with y - intercept 12.

Wait, let's re - evaluate. The function is $h(t)=12-0.25t$. So it's a linear function with slope - 0.25 and y - intercept 12. So the graph should start at (0,12) and decrease with a slope of - 0.25. Among the given graphs, the second graph (the middle top one) has a y - intercept of 12, so it's the most probable.

Answer:

The Middle - Top Graph (the second graph in the top row)