Understanding the Physics Problem: Distance Between Falling Bodies

This question asks about the relative distance between two bodies falling under gravity. One body is dropped from rest, and the other is thrown downwards with an initial velocity. We need to find the time when the separation between them reaches 18 meters.

Kinematics of Falling Bodies

We can use the equations of motion under constant acceleration to solve this problem. The acceleration acting on both bodies is the acceleration due to gravity, denoted by \(g\). Since both are moving downwards, we can take the downward direction as positive.

The general equation for displacement (\(s\)) is:

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

where:

• \(u\) is the initial velocity
• \(t\) is the time
• \(a\) is the acceleration

In this case, the acceleration \(a\) is the acceleration due to gravity \(g\).

Analyzing the Motion of Each Body

Let's consider the two bodies separately:

Body 1: Dropped from Rest

• Initial velocity, \(u_1 = 0\) m/s
• Acceleration, \(a_1 = g\)
• Distance covered in time \(t\), \(s_1\)

Using the equation of motion:

\(s_1 = u_1t + \frac{1}{2}a_1t^2\)

\(s_1 = (0)t + \frac{1}{2}gt^2\)

\(s_1 = \frac{1}{2}gt^2\)

Body 2: Thrown Downwards

• Initial velocity, \(u_2 = 1\) m/s
• Acceleration, \(a_2 = g\)
• Distance covered in time \(t\), \(s_2\)

Using the equation of motion:

\(s_2 = u_2t + \frac{1}{2}a_2t^2\)

\(s_2 = (1)t + \frac{1}{2}gt^2\)

\(s_2 = t + \frac{1}{2}gt^2\)

Calculating the Distance Between the Bodies

Both bodies start at the same instant and move downwards. Since Body 2 has an initial downward velocity while Body 1 starts from rest, Body 2 will always be ahead of Body 1. The distance between them at time \(t\) is the difference between the distances they have covered:

Distance between bodies = \(s_2 - s_1\)

We are given that the distance between them is 18 m. So,

\(s_2 - s_1 = 18\)

Substitute the expressions for \(s_1\) and \(s_2\):

\((t + \frac{1}{2}gt^2) - (\frac{1}{2}gt^2) = 18\)

Solving for Time \(t\)

Let's simplify the equation:

\(t + \frac{1}{2}gt^2 - \frac{1}{2}gt^2 = 18\)

The \(\frac{1}{2}gt^2\) terms cancel out:

\(t = 18\)

The time when the distance between the two bodies is 18 m is 18 seconds. This result is independent of the value of the acceleration due to gravity \(g\).

Conclusion

The calculation shows that the distance between the body dropped from rest and the body thrown downwards with an initial velocity of 1 m/s will be 18 m after 18 seconds.

Comparing with Options

Let's check the options provided:

1. 3 s
2. 1.8 s
3. 18 s
4. 12 s

Our calculated time, 18 seconds, matches option 3.

Revision Table: Key Concepts in Free Fall

Additional Information: Relative Velocity and Distance

This problem can also be thought of in terms of relative motion. Consider the motion of the first body (dropped from rest) as the reference frame. From the perspective of the first body, the second body has an initial relative velocity. The relative velocity of Body 2 with respect to Body 1 (\(v_{21}\)) at time \(t\) is given by \(v_{2} - v_{1}\).

• \(v_1 = u_1 + gt = 0 + gt = gt\)
• \(v_2 = u_2 + gt = 1 + gt\)

Relative velocity \(v_{21} = (1 + gt) - (gt) = 1\) m/s.

This means that the second body always moves away from the first body with a constant relative velocity of 1 m/s. This is because the acceleration due to gravity affects both bodies equally, so their relative acceleration is \(g - g = 0\).

If the relative velocity is constant at 1 m/s, the relative distance covered in time \(t\) is simply (relative velocity) \(\times\) \(t\).

Relative distance = \(1 \times t\)

We are given that the relative distance is 18 m. So,

\(1 \times t = 18\)

\(t = 18\)

This confirms the result obtained earlier using individual displacements. The key insight here is that the initial difference in velocity (1 m/s) is maintained as a constant relative velocity because both bodies experience the same gravitational acceleration.