Question

lara made the table below of the predicted values for h(t), the height, in meters, of a penny t seconds after it is dropped off of the back of the bleachers. height of penny over time t h(t) 0 2 0.1 1.951 0.2 1.804 0.3 1.559 0.4 1.216 0.5 0.775 0.6 0.236 0.7 -0.401 0.8 -1.136 to the nearest tenth of a second, how much time would it take the penny to hit the ground? 0.5 seconds 0.6 seconds 0.7 seconds 0.8 seconds

Explanation:

Step1: Understand the problem

We need to find the time \( t \) when the height \( h(t) = 0 \) (since the penny hits the ground when height is 0). We look at the table of \( t \) and \( h(t) \) values.

Step2: Analyze the table values

- At \( t = 0.6 \) seconds, \( h(t)=0.236 \) meters (positive, so penny is still above ground).
- At \( t = 0.7 \) seconds, \( h(t)= - 0.401 \) meters (negative, so penny is below ground, meaning it has hit the ground between \( t = 0.6 \) and \( t = 0.7 \)).

Step3: Interpolate (or estimate) the time

We can use linear approximation between \( t = 0.6 \) ( \( h = 0.236 \)) and \( t = 0.7 \) ( \( h=- 0.401 \)). The change in \( t \) is \( \Delta t=0.7 - 0.6 = 0.1 \) seconds, and the change in \( h \) is \( \Delta h=-0.401 - 0.236=-0.637 \) meters. We need to find \( \Delta t' \) such that \( 0 - 0.236=\frac{- 0.637}{0.1}\times\Delta t' \). Solving for \( \Delta t' \): \( \Delta t'=\frac{0.236\times0.1}{0.637}\approx0.037 \) seconds. Then the time is \( t = 0.6+\Delta t'\approx0.6 + 0.037\approx0.6 \) (wait, no, wait: actually, when \( h(t) = 0 \), we can see that at \( t = 0.6 \), \( h = 0.236 \); at \( t = 0.7 \), \( h=-0.401 \). The value of \( t \) when \( h(t) = 0 \) is closer to \( t = 0.6 \) or \( t = 0.7 \)? Let's calculate the fraction: \( \frac{0 - 0.236}{-0.637}=\frac{0.236}{0.637}\approx0.37 \). So the time is \( t=0.6 + 0.37\times0.1\approx0.6 + 0.037\approx0.64 \), but we need to round to the nearest tenth. The nearest tenth of a second between 0.6 and 0.7: 0.64 is closer to 0.6 or 0.7? Wait, 0.64 - 0.6 = 0.04, 0.7 - 0.64 = 0.06. Wait, no, wait, maybe my linear approximation is off. Alternatively, since at \( t = 0.6 \), height is 0.236 (almost 0), and at \( t = 0.7 \), it's negative. So the time when \( h(t)=0 \) is between 0.6 and 0.7. To the nearest tenth, we look at the tenths place. The value is between 0.6 and 0.7, and since 0.236 is closer to 0 than - 0.401 is to 0? Wait, no, the height at \( t = 0.6 \) is 0.236, at \( t = 0.7 \) is - 0.401. The zero crossing is closer to \( t = 0.6 \) because 0.236 is closer to 0 than 0.401 is (since distance from 0 for 0.236 is 0.236, for - 0.401 is 0.401). So using the formula for linear interpolation: \( t=t_1+\frac{0 - h(t_1)}{h(t_2)-h(t_1)}\times(t_2 - t_1) \), where \( t_1 = 0.6 \), \( h(t_1)=0.236 \), \( t_2 = 0.7 \), \( h(t_2)=-0.401 \).

So \( t = 0.6+\frac{0 - 0.236}{-0.401 - 0.236}\times(0.7 - 0.6) \)

\( t = 0.6+\frac{- 0.236}{-0.637}\times0.1 \)

\( t = 0.6+\frac{0.236}{0.637}\times0.1 \)

\( \frac{0.236}{0.637}\approx0.37 \), so \( 0.37\times0.1 = 0.037 \), so \( t\approx0.6 + 0.037=0.637 \), which is approximately 0.6 when rounded to the nearest tenth? Wait, no, 0.637 rounded to the nearest tenth is 0.6? Wait, no: the tenths place is the first decimal. 0.637: the digit in the tenths place is 6, the next digit is 3, which is less than 5? Wait, no, 0.637 is 0.6 when rounded to the nearest tenth? Wait, no, 0.637: the tenths digit is 6, the hundredths digit is 3, so we look at the hundredths digit. Since 3 < 5, we round down? Wait, no, wait: 0.637 is closer to 0.6 or 0.7? The distance from 0.6 is 0.037, from 0.7 is 0.063. So it's closer to 0.6? But wait, when \( t = 0.6 \), height is 0.236 (above ground), at \( t = 0.7 \), height is - 0.401 (below ground). So the time when it hits the ground is between 0.6 and 0.7. But let's check the options: the options are 0.5, 0.6, 0.7, 0.8. Since at \( t = 0.6 \), height is positive, at \( t = 0.7 \), height is negative. So the time is between 0.6 and 0.7. But when we round to the nearest tenth, 0.637 is approximately 0.6? Wait, no, 0.637: the tenths place is 6, the next digit is 3, so we round to 0.6? But wait, maybe I made a mistake. Wait, let's check the values again. At \( t = 0.6 \), h = 0.236 (still up). At \( t = 0.7 \), h = - 0.401 (down). So the zero crossing is between 0.6 and 0.7. The midpoint between 0.6 and 0.7 is 0.65. 0.637 is less than 0.65, so closer to 0.6? But wait, the height at 0.6 is 0.236, at 0.7 is - 0.401. The rate of change of height with respect to time (approximate velocity) is \( \frac{-0.401 - 0.236}{0.7 - 0.6}=\frac{-0.637}{0.1}=- 6.37 \) m/s. So to go from 0.236 m to 0 m, the time needed is \( \frac{0 - 0.236}{-6.37}\approx0.037 \) seconds. So total time is \( 0.6+0.037 = 0.637 \), which is approximately 0.6 when rounded to the nearest tenth? Wait, no, 0.637 rounded to the nearest tenth: the tenths digit is 6, the hundredths digit is 3, so we round down, so 0.6. But wait, let's check the options. The options are 0.5, 0.6, 0.7, 0.8. At \( t = 0.6 \), height is 0.236 (not zero), at \( t = 0.7 \), height is - 0.401 (negative, so already hit the ground). So the time when it hits the ground is between 0.6 and 0.7. But when we round to the nearest tenth, 0.637 is closer to 0.6 or 0.7? The difference between 0.637 and 0.6 is 0.037, between 0.637 and 0.7 is 0.063. So it's closer to 0.6. But wait, maybe the problem is designed so that we see that at \( t = 0.6 \), h is positive, at \( t = 0.7 \), h is negative, so the time is between 0.6 and 0.7, and since 0.236 is closer to 0 than 0.401 is (in absolute value), the nearest tenth is 0.6? Wait, no, maybe I messed up. Wait, let's think again. The height at \( t = 0.6 \) is 0.236, which is 0.236 meters above ground. At \( t = 0.7 \), it's 0.401 meters below ground. So the time when it hits the ground is when \( h(t) = 0 \), which is between 0.6 and 0.7. The formula for linear interpolation is:

Let \( t_1 = 0.6 \), \( h_1 = 0.236 \); \( t_2 = 0.7 \), \( h_2=-0.401 \)

We want to find \( t \) where \( h(t) = 0 \)

\( t=t_1+\frac{0 - h_1}{h_2 - h_1}\times(t_2 - t_1) \)

\( t = 0.6+\frac{- 0.236}{-0.401 - 0.236}\times0.1 \)

\( t = 0.6+\frac{0.236}{0.637}\times0.1 \)

\( \frac{0.236}{0.637}\approx0.37 \), so \( t\approx0.6 + 0.037 = 0.637 \), which is approximately 0.6 when rounded to the nearest tenth? Wait, 0.637 rounded to the nearest tenth: the tenths place is 6, the next digit is 3, so we round down to 0.6. But let's check the options. The options include 0.6, so that's the answer.

Answer:

0.6 seconds (B. 0.6 seconds)