Understanding Free Fall and Time Calculation

This question asks us to determine the time it takes for a freely falling body to cover half of its total fall height, given the total time taken to reach the ground from the full height.

When a body is dropped freely, its initial velocity is zero, and it moves under the constant acceleration due to gravity (g). We can use the equations of motion to solve this problem.

Applying Equations of Motion

The relevant equation of motion relating distance, initial velocity, acceleration, and time is:

\(s = ut + \frac{1}{2}at^2\)

Where:

• \(s\) is the distance covered
• \(u\) is the initial velocity
• \(a\) is the acceleration
• \(t\) is the time taken

For a body dropped freely:

• Initial velocity \(u = 0\)
• Acceleration \(a = g\) (taking downward direction as positive)

The equation simplifies to:

\(s = \frac{1}{2}gt^2\)

Step-by-Step Calculation

Step 1: Analyze the total fall time.

The body falls from a height 'X' m and takes Y seconds to reach the ground. Using the simplified equation:

\(X = \frac{1}{2}gY^2\)

We can rearrange this to find \(\frac{1}{2}g\):

\(\frac{1}{2}g = \frac{X}{Y^2}\) (Equation 1)

Step 2: Analyze the time to reach half the height.

Let \(t'\) be the time taken to reach half the height, which is \(\frac{X}{2}\) m. Using the same equation of motion:

\(\frac{X}{2} = \frac{1}{2}g(t')^2\)

\(X = g(t')^2\) (Equation 2)

Step 3: Relate the time to reach half height (\(t'\)) and the total time (Y).

We have two expressions involving X and g. Substitute the value of g from Equation 1 into Equation 2.

From Equation 1, \(\frac{1}{2}g = \frac{X}{Y^2}\), so \(g = \frac{2X}{Y^2}\).

Substitute this into Equation 2:

\(X = \left(\frac{2X}{Y^2}\right)(t')^2\)

Now, we can cancel X from both sides (assuming X > 0):

\(1 = \frac{2}{Y^2}(t')^2\)

Rearrange the equation to solve for \((t')^2\):

\((t')^2 = \frac{Y^2}{2}\)

Step 4: Solve for \(t'\).

Take the square root of both sides to find \(t'\):

\(t' = \sqrt{\frac{Y^2}{2}}\)

\(t' = \frac{\sqrt{Y^2}}{\sqrt{2}}\)

\(t' = \frac{Y}{\sqrt{2}}\)

So, the time it takes to reach half the height of the tower is \(\frac{Y}{\sqrt{2}}\) seconds.

Verification

Let's check if this result makes sense. Since the body accelerates as it falls, it covers more distance in the later part of the fall. Therefore, the time taken to cover the first half of the height should be less than half of the total time. \(\frac{Y}{\sqrt{2}}\) is indeed less than \(\frac{Y}{2}\) because \(\sqrt{2} \approx 1.414 > 1\).

Summary of Results

Solving the equations leads to \(t' = \frac{Y}{\sqrt{2}}\).

Revision Table: Free Fall Concepts

Additional Information: Relating Time and Distance in Free Fall

For an object starting from rest (\(u=0\)), the distance fallen \(s\) is proportional to the square of the time taken \(t\) (\(s \propto t^2\)). This means if you double the time, the distance fallen becomes four times; if you triple the time, the distance becomes nine times, and so on.

Conversely, the time taken to fall a certain distance is proportional to the square root of the distance (\(t \propto \sqrt{s}\)).

In this problem, the distance is reduced by half (from X to X/2). So, the new distance is \(s_{new} = \frac{1}{2}s_{original}\). The ratio of the new time \(t'\) to the original time \(Y\) is:

\(\frac{t'}{Y} = \frac{\sqrt{s_{new}}}{\sqrt{s_{original}}} = \frac{\sqrt{X/2}}{\sqrt{X}} = \sqrt{\frac{X/2}{X}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\)

Therefore, \(t' = \frac{Y}{\sqrt{2}}\). This confirms our result using the proportionality concept.