Understanding the Relationship Between Fall Time and Distance

The question describes a scenario where the time it takes for a ball to fall is related to the distance it falls. We are told that the time varies as the square root of the distance. This is a common relationship encountered in physics, particularly in problems involving constant acceleration like free fall under gravity (assuming negligible air resistance).

Mathematically, this relationship can be written as:

\( t \propto \sqrt{d} \)

Where:

• \( t \) represents the time taken to fall.
• \( d \) represents the distance fallen.
• \( \propto \) means "is proportional to".

To turn this proportionality into an equation, we introduce a constant of proportionality, let's call it \( k \):

\( t = k\sqrt{d} \)

We are given one specific case: it takes 4 seconds to fall 78.40 m. We can use this information to find the value of the constant \( k \).

\( 4 \text{ s} = k \times \sqrt{78.40 \text{ m}} \)

We can also use the fact that the ratio \( \frac{t}{\sqrt{d}} \) is constant for any distance fallen. So, if \( t_1 \) is the time to fall distance \( d_1 \), and \( t_2 \) is the time to fall distance \( d_2 \), we have:

\( \frac{t_1}{\sqrt{d_1}} = \frac{t_2}{\sqrt{d_2}} \)

This ratio method is often simpler as it avoids calculating the constant \( k \) explicitly.

Applying the Ratio Method to Solve for Fall Time

We are given the following values:

• Time \( t_1 = 4 \) seconds
• Distance \( d_1 = 78.40 \) meters
• New distance \( d_2 = 122.50 \) meters

We need to find the new time \( t_2 \) to fall the distance \( d_2 \).

Using the ratio equation:

\( \frac{t_1}{\sqrt{d_1}} = \frac{t_2}{\sqrt{d_2}} \)

Substitute the known values:

\( \frac{4 \text{ s}}{\sqrt{78.40 \text{ m}}} = \frac{t_2}{\sqrt{122.50 \text{ m}}} \)

Now, we need to solve for \( t_2 \). We can rearrange the equation:

\( t_2 = 4 \text{ s} \times \frac{\sqrt{122.50 \text{ m}}}{\sqrt{78.40 \text{ m}}} \)

We can simplify the square roots by combining them:

\( t_2 = 4 \text{ s} \times \sqrt{\frac{122.50 \text{ m}}{78.40 \text{ m}}} \)

Step-by-Step Calculation

Let's perform the calculation step-by-step:

1. Calculate the ratio of the distances inside the square root:

   \( \frac{122.50}{78.40} \)

   \( 122.50 \div 78.40 = 1.5625 \)

2. Take the square root of this ratio:

   \( \sqrt{1.5625} \)

   \( \sqrt{1.5625} = 1.25 \)

3. Multiply this result by the initial time \( t_1 \):

   \( t_2 = 4 \text{ s} \times 1.25 \)

   \( t_2 = 5 \text{ s} \)

So, it would take 5 seconds for the ball to fall 122.50 m.

Revision Table: Key Concepts for Fall Time Calculation

Additional Information on Free Fall and Time

The relationship \( t \propto \sqrt{d} \) is specifically true for objects falling under constant acceleration from rest, like in free fall near the Earth's surface (ignoring air resistance). The actual formula relating distance and time in free fall from rest is \( d = \frac{1}{2}gt^2 \), where \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).

From \( d = \frac{1}{2}gt^2 \), we can rearrange to solve for \( t \):

\( t^2 = \frac{2d}{g} \)

\( t = \sqrt{\frac{2d}{g}} \)

\( t = \sqrt{\frac{2}{g}} \times \sqrt{d} \)

Comparing this to \( t = k\sqrt{d} \), we see that the constant of proportionality \( k \) is equal to \( \sqrt{\frac{2}{g}} \). Using \( g = 9.8 \text{ m/s}^2 \):

\( k = \sqrt{\frac{2}{9.8}} = \sqrt{\frac{1}{4.9}} \approx \sqrt{0.204} \approx 0.4517 \)

This matches the value of \( k \) we would have calculated using the first method \( k = \frac{4}{\sqrt{78.40}} \approx \frac{4}{8.854} \approx 0.4517 \). This confirms the stated relationship is consistent with the physics of free fall.