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

$t$	$h(t)$
0.1	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?

• 0.6 seconds
• 0.5 seconds
• 0.7 seconds
• 0.8 seconds

Explanation:

Step1: Understand the problem

We need to find when the penny hits the ground, which means \( h(t) = 0 \) (height is 0 meters). We look at the table of \( t \) (time in seconds) and \( h(t) \) (height in meters) to see when \( h(t) \) is closest to 0, and then determine the time \( t \) to the nearest tenth of a second.

Step2: Analyze the table values

- At \( t = 0.5 \) seconds, \( h(t)=0.775 \) meters (positive, above ground).
- At \( t = 0.6 \) seconds, \( h(t)=0.236 \) meters (positive, above ground but closer to 0 than at \( t = 0.5 \)).
- At \( t = 0.7 \) seconds, \( h(t)= - 0.401 \) meters (negative, below ground, meaning it has hit the ground before or at this time).
- At \( t = 0.8 \) seconds, \( h(t)= - 1.136 \) meters (negative, further below ground).

We need to see between which two times the height crosses from positive to negative (since that's when it hits the ground). Between \( t = 0.6 \) ( \( h = 0.236 \)) and \( t = 0.7 \) ( \( h=- 0.401 \) ), the height goes from positive to negative. Now we check which time is closer to when \( h(t) = 0 \).

The value at \( t = 0.6 \) is \( 0.236 \) (distance from 0 is \( 0.236 \)), and at \( t = 0.7 \) is \( 0.401 \) (distance from 0 is \( 0.401 \)). Wait, no, actually, when moving from \( t = 0.6 \) ( \( h = 0.236 \)) to \( t = 0.7 \) ( \( h=-0.401 \) ), the zero crossing is between 0.6 and 0.7. But we need to find to the nearest tenth. Let's calculate the difference:

At \( t = 0.6 \), \( h = 0.236 \); at \( t = 0.7 \), \( h=-0.401 \). The change in \( h \) is \( - 0.401-0.236=-0.637 \) over a time change of \( 0.7 - 0.6 = 0.1 \) seconds. We need to find \( \Delta t \) such that \( 0.236+\frac{0 - 0.236}{- 0.637}\times0.1 \). Wait, maybe simpler: look at the values. At \( t = 0.6 \), height is 0.236 (still 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. Now, which is closer to 0.6 or 0.7? The height at 0.6 is 0.236, at 0.7 is - 0.401. The absolute value of 0.236 is 0.236, and of - 0.401 is 0.401. Wait, no, actually, when the height goes from positive to negative, the zero crossing is between 0.6 and 0.7. But the question is to the nearest tenth. Let's check the options. The options are 0.6, 0.5, 0.7, 0.8.

Wait, maybe I made a mistake. Wait, at \( t = 0.6 \), \( h = 0.236 \) (above ground), at \( t = 0.7 \), \( h=-0.401 \) (below ground). So the penny hits the ground between 0.6 and 0.7 seconds. Now, to find the nearest tenth, we see that 0.6 is 0.236 above ground, 0.7 is 0.401 below ground. The distance from 0.6 to 0 (in terms of height) is 0.236, and from 0.7 to 0 (in terms of height) is 0.401. Wait, no, the time when it hits the ground is when \( h(t) = 0 \). Let's use linear approximation between \( t = 0.6 \) ( \( h = 0.236 \)) and \( t = 0.7 \) ( \( h=-0.401 \) ).

The formula for linear approximation is \( t = t_1+\frac{0 - h(t_1)}{h(t_2)-h(t_1)}\times(t_2 - t_1) \), where \( t_1 = 0.6 \), \( h(t_1)=0.236 \), \( t_2 = 0.7 \), \( h(t_2)=-0.401 \).

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

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

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

\( \frac{0.236}{0.637}\approx0.370 \)

\( t\approx0.6 + 0.037\approx0.637 \) seconds. Rounding to the nearest tenth, that's 0.6 seconds? Wait, no, wait 0.637 is closer to 0.6 or 0.7? 0.637 - 0.6 = 0.037, 0.7 - 0.637 = 0.063. So 0.637 is closer to 0.6 (since 0.037 < 0.063)? Wait, no, 0.637 is 0.6 when rounded to the nearest tenth? Wait, 0.637: the tenths place is 6, the hundredths place is 3, which is less than 5, so we round down? Wait, no, wait 0.637 is 0.6 when rounded to the nearest tenth? Wait, no, 0.637: the first decimal is 6 (tenths), second is 3 (hundredths). Since 3 < 5, we keep the tenths place as 6. But wait, the height at 0.6 is 0.236 (above ground), at 0.7 is - 0.401 (below ground). So the time when it hits the ground is between 0.6 and 0.7. But the question is to the nearest tenth. Let's check the options again. The options are 0.6, 0.5, 0.7, 0.8.

Wait, maybe I messed up. Let's look at the values again:

At t=0.5: 0.775 (above)

t=0.6: 0.236 (above)

t=0.7: -0.401 (below)

So between 0.6 and 0.7, it hits the ground. Now, which is closer to 0.6 or 0.7?

The difference between 0.236 (at 0.6) and 0 is 0.236.

The difference between -0.401 (at 0.7) and 0 is 0.401.

Wait, no, the time when h(t)=0 is between 0.6 and 0.7. Let's calculate the fraction:

The change in h from 0.6 to 0.7 is -0.401 - 0.236 = -0.637.

We need to find how much time after 0.6 seconds does it take to reach h=0.

So the fraction is (0 - 0.236)/(-0.637) = 0.236/0.637 ≈ 0.37.

So the time is 0.6 + 0.37*0.1 ≈ 0.6 + 0.037 ≈ 0.637 seconds, which is approximately 0.6 seconds when rounded to the nearest tenth? Wait, no, 0.637 rounded to the nearest tenth is 0.6? Wait, 0.637: the tenths digit is 6, the hundredths digit is 3, which is less than 5, so we round down to 0.6? But wait, 0.637 is closer to 0.6 than to 0.7? Wait, 0.7 - 0.637 = 0.063, 0.637 - 0.6 = 0.037. Yes, 0.037 < 0.063, so it's closer to 0.6. But wait, the height at 0.6 is 0.236 (still above), at 0.7 is -0.401 (below). So the time when it hits the ground is 0.6 + (0 - 0.236)/(-0.637) * 0.1 ≈ 0.637, which is 0.6 when rounded to the nearest tenth? But wait, the options include 0.6, 0.7, etc. Wait, maybe the question is considering that at 0.6 seconds, it's still above ground (0.236 m), at 0.7 seconds, it's below ground (-0.401 m). So the time to hit the ground is between 0.6 and 0.7, and to the nearest tenth, since 0.637 is closer to 0.6? Wait, no, 0.637 is 0.6 when rounded to the nearest tenth? Wait, 0.637: the first decimal is 6, second is 3, so we round to 0.6? But let's check the options. The options are 0.6, 0.5, 0.7, 0.8.

Wait, maybe I made a mistake in the calculation. Let's think differently. The height at t=0.6 is 0.236 (positive), at t=0.7 is -0.401 (negative). So the penny hits the ground between 0.6 and 0.7 seconds. Now, which is closer to 0.6 or 0.7?

The value 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 0.6 + (0.236 / (0.236 + 0.401)) * 0.1.

Because the total distance from 0.236 (above) to -0.401 (below) is 0.236 + 0.401 = 0.637 meters.

The distance from 0.236 to 0 is 0.236, so the fraction is 0.236 / 0.637 ≈ 0.37.

So 0.6 + 0.37*0.1 ≈ 0.637, which is 0.6 when rounded to the nearest tenth (since 0.637 is 0.6 when rounded to one decimal place: the second decimal is 3, which is less than 5, so we keep the first decimal as 6).

Wait, but let's check the options. The options are 0.6, 0.5, 0.7, 0.8. So the correct answer should be 0.6 seconds? Wait, no, wait 0.637 is closer to 0.6 than to 0.7? Wait, 0.637 is 0.6 when rounded to the nearest tenth? Wait, 0.637: the tenths place is 6, hundredths is 3, so yes, round to 0.6. But wait, maybe the problem is that at t=0.6, it's still above, at t=0.7, below. So the time is between 0.6 and 0.7, and to the nearest tenth, it's 0.6? But let's check the answer options. The first option is 0.6 seconds.

Wait, maybe I was overcomplicating. Let's see the values:

t=0.5: 0.775

t=0.6: 0.236

t=0.7: -0.401

t=0.8: -1.136

So the height goes from positive to negative between 0.6 and 0.7. Now, which is closer to 0.6 or 0.7?

The height at 0.6 is 0.236 (close to 0), at 0.7 is -0.401 (further from 0). So the time when it hits the ground is closer to 0.6 seconds. So the answer is 0.6 seconds.

Answer:

0.6 seconds