We are tasked with finding the relative error in the determination of time $t$ for a ball dropped from height $H$ using the formula:

$t = \sqrt{\frac{2H}{g}}$

where:

• $H = 1.490 \, \text{m}$ with a probable error of $0.018 \, \text{m}$,
• $g = 9.82 \, \text{m/s}^2$ with a probable error of $0.44 \, \text{m/s}^2$.

Step 1: Expression for the relative error of $t$

To find the relative error in $t$, we first need to differentiate $t = \sqrt{\frac{2H}{g}}$ with respect to both $H$ and $g$, treating the errors as small differentials:

$\frac{\Delta t}{t} = \frac{1}{2} \left( \frac{\Delta H}{H} + \frac{\Delta g}{g} \right)$

Step 2: Calculate the relative errors for $H$ and $g$

• Relative error in height $H$: $\frac{\Delta H}{H} = \frac{0.018}{1.490}$

• Relative error in gravitational acceleration $g$: $\frac{\Delta g}{g} = \frac{0.44}{9.82}$

Step 3: Combine the relative errors

Using the formula for the relative error in $t$, we substitute the relative errors from above:

$\frac{\Delta t}{t} = \frac{1}{2} \left( \frac{0.018}{1.490} + \frac{0.44}{9.82} \right)$

Now, let me calculate the final result.The relative errors for height $H$ and gravitational acceleration $g$ are:

• Relative error in $H$: $0.0121$ (or 1.21%)
• Relative error in $g$: $0.0448$ (or 4.48%)

Thus, the relative error in the time $t$ is approximately $0.0284$, or 2.84%.

Would you like more details on how to propagate errors, or do you have any questions about the steps?

Here are some related questions:

1. How would the time $t$ change if the height $H$ was doubled?
2. How would the relative error in time change if the error in $H$ was reduced by half?
3. How does the uncertainty in $g$ affect the overall error more than the uncertainty in $H$?
4. What assumptions are made when calculating relative errors for small differentials?
5. How would you account for systematic errors in a real experiment when measuring $H$?

Tip: Always express uncertainties with the same precision as the measured value to maintain consistency in reporting data.