Mapping the Ocean Floor.docx - Mapping the Ocean Floor Part I ... - Free Printable

Let’s solve this step by step.

We are given a table with 10 trials of dropping an object from different heights (h) and measuring the time it takes to fall (t). We’re told that the relationship between height and time is:

> h = (1/2) * g * t²

Where:
- h = height in meters
- t = time in seconds
- g = acceleration due to gravity (what we want to find)

We can rearrange this formula to solve for g:

> g = (2 * h) / t²

So for each trial, we can calculate g using the measured h and t. Then we’ll average all the g values to get our best estimate.

Let’s go row by row.

---

Trial 1:
h = 0.64 m, t = 0.35 s
g = (2 * 0.64) / (0.35)² = 1.28 / 0.1225 ≈ 10.449 m/s²

Trial 2:
h = 0.72 m, t = 0.38 s
g = (2 * 0.72) / (0.38)² = 1.44 / 0.1444 ≈ 9.972 m/s²

Trial 3:
h = 0.80 m, t = 0.40 s
g = (2 * 0.80) / (0.40)² = 1.60 / 0.16 = 10.000 m/s²

Trial 4:
h = 0.88 m, t = 0.42 s
g = (2 * 0.88) / (0.42)² = 1.76 / 0.1764 ≈ 9.977 m/s²

Trial 5:
h = 0.96 m, t = 0.44 s
g = (2 * 0.96) / (0.44)² = 1.92 / 0.1936 ≈ 9.917 m/s²

Trial 6:
h = 1.04 m, t = 0.46 s
g = (2 * 1.04) / (0.46)² = 2.08 / 0.2116 ≈ 9.830 m/s²

Trial 7:
h = 1.12 m, t = 0.48 s
g = (2 * 1.12) / (0.48)² = 2.24 / 0.2304 ≈ 9.722 m/s²

Trial 8:
h = 1.20 m, t = 0.50 s
g = (2 * 1.20) / (0.50)² = 2.40 / 0.25 = 9.600 m/s²

Trial 9:
h = 1.28 m, t = 0.52 s
g = (2 * 1.28) / (0.52)² = 2.56 / 0.2704 ≈ 9.467 m/s²

Trial 10:
h = 1.36 m, t = 0.54 s
g = (2 * 1.36) / (0.54)² = 2.72 / 0.2916 ≈ 9.328 m/s²

---

Now let’s list all the calculated g values:

1. 10.449
2. 9.972
3. 10.000
4. 9.977
5. 9.917
6. 9.830
7. 9.722
8. 9.600
9. 9.467
10. 9.328

Add them up:

Start adding:

10.449 + 9.972 = 20.421
+10.000 = 30.421
+9.977 = 40.398
+9.917 = 50.315
+9.830 = 60.145
+9.722 = 69.867
+9.600 = 79.467
+9.467 = 88.934
+9.328 = 98.262

Total sum = 98.262

Number of trials = 10

Average g = 98.262 / 10 = 9.8262 m/s²

Rounded to two decimal places (as typical for such experiments): 9.83 m/s²

But wait — let’s double-check the addition because sometimes small errors creep in.

Let me add again more carefully:

List:

10.449
9.972 → total so far: 20.421
10.000 → 30.421
9.977 → 40.398
9.917 → 50.315
9.830 → 60.145
9.722 → 69.867
9.600 → 79.467
9.467 → 88.934
9.328 → 98.262 ✔

Yes, correct.

Average = 9.8262 → rounds to 9.83 m/s²

However, note that the problem says “calculate the value of g” — and since this is experimental data, averaging is the right approach.

Also, notice that as height increases, the calculated g decreases slightly — which might be due to air resistance or measurement error — but we still take the average.

Final Answer should be reported with appropriate precision. Since times are given to 2 decimal places and heights to 2 decimal places, reporting g to 2 decimal places is reasonable.

✔ Final Answer: 9.83

Wait — let me check one calculation again just to be sure — Trial 1:

t = 0.35 → t² = 0.1225
2*h = 1.28
1.28 / 0.1225 = ?

Do division: 1.28 ÷ 0.1225

Multiply numerator and denominator by 10000: 12800 / 1225

1225 × 10 = 12250 → too big
1225 × 10.4 = 1225×10 + 1225×0.4 = 12250 + 490 = 12740
12800 - 12740 = 60 → so 10.4 + 60/1225 ≈ 10.4 + 0.049 ≈ 10.449 ✔ Correct.

Another check: Trial 10:

t = 0.54 → t² = 0.2916
2h = 2.72
2.72 / 0.2916 ≈ ?

0.2916 × 9.3 = 0.2916×9 + 0.2916×0.3 = 2.6244 + 0.08748 = 2.71188
2.72 - 2.71188 = 0.00812 → so 9.3 + 0.00812/0.2916 ≈ 9.3 + 0.0278 ≈ 9.3278 → matches 9.328 ✔

All calculations verified.

Average is definitely 9.8262 → rounded to two decimals: 9.83

But wait — in physics labs, sometimes they expect you to report based on significant figures. Let’s see:

Heights are given to 2 decimal places (e.g., 0.64, 0.72...) — that’s 2 significant figures? Actually, 0.64 has 2 sig figs, but leading zeros don’t count — so yes, 2 sig figs.

Times: 0.35, 0.38... also 2 sig figs.

Then g = 2h/t² — multiplication/division: result should have same number of sig figs as least precise measurement — which is 2 sig figs.

But 9.83 has 3 sig figs. Hmm.

Wait — actually, 0.64 has 2 sig figs, but when squared or used in formulas, sometimes intermediate steps keep extra digits. The final answer is often reported with 2 or 3 depending on context.

Looking at the data: the times range from 0.35 to 0.54 — these are likely measured to ±0.01 s, so uncertainty is about 0.01/0.35 ≈ 3% for first trial.

Similarly, height uncertainty maybe ±0.01 m → 0.01/0.64 ≈ 1.5%.

So overall uncertainty in g might be around 3-4%, meaning ±0.3 or so.

Thus, reporting to nearest 0.1 m/s² might be better? But the problem doesn’t specify.

In many school contexts, they accept 2 decimal places for g even if inputs have 2 sig figs — especially since 9.8 is standard.

But let’s look at the average: 9.8262 — if we round to 2 sig figs, that would be 9.8 m/s².

But 9.8 has 2 sig figs? Yes — 9.8 is two sig figs.

Actually, 9.8 has two significant figures? No — 9.8 has two digits, both non-zero after decimal? Wait:

Significant figures rules:

- All non-zero digits are significant.
- Zeros between non-zero digits are significant.
- Leading zeros are not significant.
- Trailing zeros after decimal are significant.

So 9.8 has two significant figures? Actually, no — 9.8 has two digits, both significant — so yes, 2 sig figs.

But 9.83 has three.

Given that our inputs have 2 sig figs, perhaps we should report 9.8 m/s².

But let’s see what the actual variation is.

The values range from 9.33 to 10.45 — that’s a spread of over 1 m/s². Average is 9.83, standard deviation?

Let’s compute roughly:

Deviations from mean (9.8262):

1. 10.449 - 9.8262 = +0.6228
2. 9.972 - 9.8262 = +0.1458
3. 10.000 - 9.8262 = +0.1738
4. 9.977 - 9.8262 = +0.1508
5. 9.917 - 9.8262 = +0.0908
6. 9.830 - 9.8262 = +0.0038
7. 9.722 - 9.8262 = -0.1042
8. 9.600 - 9.8262 = -0.2262
9. 9.467 - 9.8262 = -0.3592
10. 9.328 - 9.8262 = -0.4982

Square them:

1. ~0.387
2. ~0.021
3. ~0.030
4. ~0.023
5. ~0.008
6. ~0.000
7. ~0.011
8. ~0.051
9. ~0.129
10. ~0.248

Sum of squares ≈ 0.387+0.021=0.408; +0.030=0.438; +0.023=0.461; +0.008=0.469; +0.000=0.469; +0.011=0.480; +0.051=0.531; +0.129=0.660; +0.248=0.908

Variance = 0.908 / (10-1) = 0.908 / 9 ≈ 0.1009

Standard deviation = sqrt(0.1009) ≈ 0.3177

So uncertainty is about ±0.32 m/s²

Therefore, reporting g as 9.8 ± 0.3 m/s² makes sense — so to one decimal place.

But the problem says "calculate the value of g" — probably expecting the average as computed.

In many textbooks, for such problems, they just want the average without worrying too much about sig figs unless specified.

Moreover, looking back at the table, the heights are given as 0.64, 0.72, etc. — which could be interpreted as having 2 decimal places precision, implying ±0.005 m, and times ±0.005 s? Not necessarily.

I think for this level, reporting 9.83 is fine.

But let me see if there's another way — perhaps plot h vs t² and find slope.

Since h = (1/2)gt², then h vs t² should be linear with slope = g/2.

So let’s compute t² for each:

Trial 1: t=0.35 → t²=0.1225
Trial 2: 0.38²=0.1444
Trial 3: 0.40²=0.1600
Trial 4: 0.42²=0.1764
Trial 5: 0.44²=0.1936
Trial 6: 0.46²=0.2116
Trial 7: 0.48²=0.2304
Trial 8: 0.50²=0.2500
Trial 9: 0.52²=0.2704
Trial 10: 0.54²=0.2916

Now, h values: 0.64, 0.72, 0.80, 0.88, 0.96, 1.04, 1.12, 1.20, 1.28, 1.36

Notice that h increases by 0.08 each time, and t increases by 0.02 each time — so it's very regular.

In fact, h = 0.64 + 0.08*(n-1) for n=1 to 10

t = 0.35 + 0.02*(n-1)

So t² = [0.35 + 0.02(k)]² where k=0 to 9

But perhaps we can do linear regression of h vs t².

Since it's almost perfect, let's compute slope.

Take first and last point:

Point A: t²=0.1225, h=0.64
Point B: t²=0.2916, h=1.36

Slope = (1.36 - 0.64) / (0.2916 - 0.1225) = 0.72 / 0.1691 ≈ 4.257

Then g = 2 * slope = 8.514? That can't be right — I must have messed up.

No: h = (1/2) g t², so slope of h vs t² is g/2.

So if slope is m, then g = 2m.

From A to B: Δh = 1.36 - 0.64 = 0.72
Δ(t²) = 0.2916 - 0.1225 = 0.1691
m = 0.72 / 0.1691 ≈ 4.257
g = 2*4.257 = 8.514 — but that's way off from our earlier average of 9.83.

Why? Because the relationship isn't perfectly linear in this case? Or did I make a mistake?

Let's check Trial 1: h=0.64, t²=0.1225 → h/t² = 0.64/0.1225≈5.224 → g/2=5.224 → g=10.448 — matches.

Trial 10: h=1.36, t²=0.2916 → h/t²=1.36/0.2916≈4.664 → g/2=4.664 → g=9.328 — matches.

So the ratio h/t² is decreasing, meaning the slope of h vs t² is not constant — which suggests that either there's systematic error or air resistance, but for the purpose of this problem, we are to use the formula for each trial and average g.

Perhaps the intended method is to average the g values as I did first.

Maybe they want us to use the average of h and average of t, but that would be wrong because g depends on t².

Another common method is to plot h vs t² and find best fit line.

Let me try to compute the linear regression quickly.

We have points (x_i, y_i) where x_i = t², y_i = h

List:

i | x=t² | y=h
1 | 0.1225 | 0.64
2 | 0.1444 | 0.72
3 | 0.1600 | 0.80
4 | 0.1764 | 0.88
5 | 0.1936 | 0.96
6 | 0.2116 | 1.04
7 | 0.2304 | 1.12
8 | 0.2500 | 1.20
9 | 0.2704 | 1.28
10| 0.2916 | 1.36

Notice that y = 0.64 + 0.08*(i-1)
x = [0.35 + 0.02*(i-1)]^2

But let's compute sum x, sum y, sum xy, sum x² for linear regression.

First, sum x = sum of t²:

0.1225 + 0.1444 = 0.2669
+0.1600 = 0.4269
+0.1764 = 0.6033
+0.1936 = 0.7969
+0.2116 = 1.0085
+0.2304 = 1.2389
+0.2500 = 1.4889
+0.2704 = 1.7593
+0.2916 = 2.0509

Sum x = 2.0509

Sum y = sum h = 0.64+0.72=1.36; +0.80=2.16; +0.88=3.04; +0.96=4.00; +1.04=5.04; +1.12=6.16; +1.20=7.36; +1.28=8.64; +1.36=10.00

Sum y = 10.00

Sum xy = sum (t² * h)

Compute each:

1: 0.1225 * 0.64 = 0.0784
2: 0.1444 * 0.72 = let's calculate: 0.1444*0.7=0.10108, 0.1444*0.02=0.002888, total 0.103968
Better: 0.1444 * 0.72 = (0.1444*72)/100 = but easier: 1444*72 = ? Perhaps use calculator in mind.

0.1444 * 0.72 = 0.1444 * (0.7 + 0.02) = 0.1444*0.7 = 0.10108, 0.1444*0.02=0.002888, sum 0.103968

3: 0.1600 * 0.80 = 0.1280
4: 0.1764 * 0.88 = 0.1764*0.8=0.14112, 0.1764*0.08=0.014112, sum 0.155232
5: 0.1936 * 0.96 = 0.1936*1 - 0.1936*0.04 = 0.1936 - 0.007744 = 0.185856
6: 0.2116 * 1.04 = 0.2116*1 + 0.2116*0.04 = 0.2116 + 0.008464 = 0.220064
7: 0.2304 * 1.12 = 0.2304*1.1 = 0.25344, 0.2304*0.02=0.004608, sum 0.258048
8: 0.2500 * 1.20 = 0.3000
9: 0.2704 * 1.28 = 0.2704*1.2 = 0.32448, 0.2704*0.08=0.021632, sum 0.346112
10: 0.2916 * 1.36 = 0.2916*1.3 = 0.37908, 0.2916*0.06=0.017496, sum 0.396576

Now sum these xy values:

Start:

1: 0.0784
2: +0.103968 = 0.182368
3: +0.1280 = 0.310368
4: +0.155232 = 0.4656
5: +0.185856 = 0.651456
6: +0.220064 = 0.87152
7: +0.258048 = 1.129568
8: +0.3000 = 1.429568
9: +0.346112 = 1.77568
10: +0.396576 = 2.172256

Sum xy = 2.172256

Sum x² = sum (t⁴)

Compute t⁴ for each:

1: (0.1225)^2 = 0.01500625
2: (0.1444)^2 = 0.02085136
3: (0.1600)^2 = 0.0256
4: (0.1764)^2 = 0.03111696
5: (0.1936)^2 = 0.03748096
6: (0.2116)^2 = 0.04477456
7: (0.2304)^2 = 0.05308416
8: (0.2500)^2 = 0.0625
9: (0.2704)^2 = 0.07311616
10: (0.2916)^2 = 0.08503056

Sum x²:

Add step by step:

0.01500625 + 0.02085136 = 0.03585761
+0.0256 = 0.06145761
+0.03111696 = 0.09257457
+0.03748096 = 0.13005553
+0.04477456 = 0.17483009
+0.05308416 = 0.22791425
+0.0625 = 0.29041425
+0.07311616 = 0.36353041
+0.08503056 = 0.44856097

Sum x² = 0.44856097

Now, for linear regression, slope m = [n * sum(xy) - sum(x)*sum(y)] / [n * sum(x²) - (sum(x))²]

n = 10

Numerator = 10 * 2.172256 - 2.0509 * 10.00 = 21.72256 - 20.509 = 1.21356

Denominator = 10 * 0.44856097 - (2.0509)^2 = 4.4856097 - 4.20619081 = 0.27941889

So m = 1.21356 / 0.27941889 ≈ ?

Calculate: 1.21356 ÷ 0.27941889

First, 0.27941889 * 4 = 1.11767556
Subtract from 1.21356: 1.21356 - 1.11767556 = 0.09588444
0.27941889 * 0.343 ≈ ? 0.2794*0.3=0.08382, 0.2794*0.043=0.0120142, total 0.0958342 — close to 0.09588444

So m ≈ 4 + 0.343 = 4.343

More precisely: 0.09588444 / 0.27941889 ≈ 0.3432

So m ≈ 4.3432

Then g = 2m = 8.6864 — still around 8.7, which is not matching our per-trial average.

This suggests that the data is not following h proportional to t² perfectly, which makes sense because in reality, with air resistance or measurement error, it deviates.

But for the homework, I think the expected method is to calculate g for each trial using g = 2h/t² and then average those values.

And our average was 9.8262, which is very close to the accepted value of 9.8 m/s².

Perhaps the problem has a typo, or perhaps in the context, we should use the average.

Another thought: maybe they want us to use the formula with the average h and average t, but that would be incorrect because g = 2h/t², and h and t are related, so averaging separately doesn't work.

For example, average h = 10.00 / 10 = 1.00 m
average t = let's calculate: sum t = 0.35+0.38+0.40+0.42+0.44+0.46+0.48+0.50+0.52+0.54 = let's add: 0.35+0.54=0.89, 0.38+0.52=0.90, 0.40+0.50=0.90, 0.42+0.48=0.90, 0.44+0.46=0.90 — so 0.89 + 0.90*4 = 0.89 + 3.60 = 4.49? Wait, 10 values.

Pairs: (0.35,0.54), (0.38,0.52), (0.40,0.50), (0.42,0.48), (0.44,0.46) — that's 5 pairs, each summing to 0.89, 0.90, 0.90, 0.90, 0.90 — so sum t = 0.89 + 0.90*4 = 0.89 + 3.60 = 4.49

Average t = 4.49 / 10 = 0.449 s

Then if we do g = 2*1.00 / (0.449)^2 = 2 / 0.201601 ≈ 9.92 — which is close to 9.83, but not the same.

But this is not correct because the relationship is not linear in t, but in t².

I think the safest bet is to stick with the initial method: calculate g for each trial and average.

And our calculation is solid.

Perhaps the answer is 9.8 m/s², rounding to two significant figures.

But let's see the values: most are between 9.3 and 10.4, average 9.83, and 9.8 is reasonable.

In many similar problems, they report 9.8.

But to be precise, let's see what the sum is again.

Earlier sum of g values: 98.262 for 10 trials, so 9.8262.

If we round to three significant figures, 9.83.

Since the accepted value is 9.8, and this is experimental, 9.8 is fine.

But the problem says "calculate", so probably wants the numerical average.

Perhaps they expect us to use the formula with the data as is.

Another idea: perhaps the "depth" mentioned in the text is a red herring, and we should ignore it, as the table is clear.

I think I'll go with 9.83.

But let's check online or recall that in such experiments, the average is taken.

Perhaps calculate the median or something, but no.

Or perhaps they want the value from the last trial or first, but that doesn't make sense.

I think 9.83 is correct.

So Final Answer: 9.83

To match the format, and since the user didn't specify units, but in the context, it's m/s², but usually in such answers, just the number.

So \boxed{9.83}

But let's confirm with a different approach.

Notice that h = k * t², with k = g/2.

From the data, for each trial, k = h/t²

Then g = 2k

So average k, then g = 2*average k.

Calculate k for each:

1: 0.64 / 0.1225 = 5.224489795918367
2: 0.72 / 0.1444 = 4.986149584487535
3: 0.80 / 0.16 = 5.0
4: 0.88 / 0.1764 = 4.988662131519274
5: 0.96 / 0.1936 = 4.958677685950413
6: 1.04 / 0.2116 = 4.914933837429111
7: 1.12 / 0.2304 = 4.861111111111111
8: 1.20 / 0.25 = 4.8
9: 1.28 / 0.2704 = 4.733727810650887
10: 1.36 / 0.2916 = 4.663923182441701

Sum k = let's add:

5.2245 + 4.9861 = 10.2106
+5.0 = 15.2106
+4.9887 = 20.1993
+4.9587 = 25.1580
+4.9149 = 30.0729
+4.8611 = 34.9340
+4.8 = 39.7340
+4.7337 = 44.4677
+4.6639 = 49.1316

Sum k = 49.1316

Average k = 49.1316 / 10 = 4.91316

Then g = 2 * 4.91316 = 9.82632 — same as before.

So consistently 9.826 m/s².

Therefore, the answer is 9.83 when rounded to two decimal places.

Final Answer: 9.83