Understanding Falling Body Motion

This problem involves calculating the additional distance a body needs to fall to double its velocity, starting from rest.

Physics Principles for Falling Objects

We use the kinematic equation relating final velocity (\(v_f\)), initial velocity (\(v_i\)), acceleration (\(a\)), and distance (\(d\)):

\(v_f^2 = v_i^2 + 2ad\)

Here, the acceleration is due to gravity (\(a = g\)), and the body starts from rest (\(v_i = 0\)).

Calculating Velocity After Distance '\(h\)'

Let the velocity after falling a distance '\(h\)' be '\(v\)'. Using the kinematic equation:

\(v^2 = 0^2 + 2gh\)

\(v^2 = 2gh \quad \quad (1)\)

Calculating Distance for Doubled Velocity

We want the velocity to become '\(2v\)'. Let the total distance fallen to achieve this velocity be '\(H\)'.

Using the same kinematic equation:

\((2v)^2 = 0^2 + 2gH\)

\(4v^2 = 2gH \quad \quad (2)\)

Deriving the Further Distance

Substitute the value of '\(v^2\)' from equation (1) into equation (2):

\(4(2gh) = 2gH\)

\(8gh = 2gH\)

Divide both sides by '\(2g\)':

\(H = \frac{8gh}{2g}\)

\(H = 4h\)

This '\(H\)' is the total distance fallen. The question asks for the *further* distance needed after the initial fall of '\(h\)'.

Further Distance = Total Distance (\(H\)) - Initial Distance (\(h\))

Further Distance = \(4h - h = 3h\)

Therefore, the body has to fall an additional distance of 3 times '\(h\)'.