When a ball is allowed to fall, the time it takes to fall any distance varies as the square root of the distance and it takes 4 seconds to fall 78.40 m. How long would it take to fall 122.50 m?
5 seconds
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:
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.
We are given the following values:
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:
\( t_2 = 4 \text{ s} \times \sqrt{\frac{122.50 \text{ m}}{78.40 \text{ m}}} \)
Let's perform the calculation step-by-step:
\( \frac{122.50}{78.40} \)
\( 122.50 \div 78.40 = 1.5625 \)
\( \sqrt{1.5625} \)
\( \sqrt{1.5625} = 1.25 \)
\( t_2 = 4 \text{ s} \times 1.25 \)
\( t_2 = 5 \text{ s} \)
So, it would take 5 seconds for the ball to fall 122.50 m.
| Quantity | Symbol | Value 1 | Value 2 |
|---|---|---|---|
| Time | \( t \) | \( t_1 = 4 \) s | \( t_2 = ? \) |
| Distance | \( d \) | \( d_1 = 78.40 \) m | \( d_2 = 122.50 \) m |
| Concept | Description | Formula/Relation |
|---|---|---|
| Proportionality | Two quantities vary together such that their ratio remains constant. | \( y \propto x \) or \( y = kx \) |
| Square Root Variation | A quantity varies directly as the square root of another quantity. | \( t \propto \sqrt{d} \) or \( t = k\sqrt{d} \) |
| Ratio Method | For proportional relationships, the ratio of corresponding quantities is constant. | \( \frac{y_1}{x_1} = \frac{y_2}{x_2} \) or \( \frac{t_1}{\sqrt{d_1}} = \frac{t_2}{\sqrt{d_2}} \) |
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.
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