A body freely falling from rest has acquired a velocity ‘v’ after it falls through a distance ‘h’. The distance it has to fall down further for its velocity to become double is:
3h
Understanding Free Fall Motion
When a body is freely falling, it means it is moving under the influence of gravity alone. The acceleration of the body is constant and equal to the acceleration due to gravity, denoted by \(g\). For motion downwards, we can consider \(g\) as positive if we define the downward direction as positive.
The body starts from rest, which means its initial velocity is zero.
Applying Kinematic Equations to Free Fall
We can use the standard kinematic equations for uniformly accelerated motion. The most relevant equation for this problem, which relates initial velocity (\(u\)), final velocity (\(v_{f}\)), acceleration (\(a\)), and distance (\(s\)), is:
\[v_{f}^2 = u^2 + 2as\]
In the case of free fall starting from rest, \(u = 0\) and \(a = g\). So the equation simplifies to:
\[v_{f}^2 = 0^2 + 2gs\]
\[v_{f}^2 = 2gs\]
Step-by-Step Calculation
Part 1: Falling Distance 'h' to Reach Velocity 'v'
The body starts from rest (\(u_1 = 0\)) and falls a distance \(s_1 = h\), reaching a velocity \(v_{f1} = v\).
Using the kinematic equation \(v_{f}^2 = 2gs\):
\[v^2 = 2gh \quad \cdots (1)\]
This equation relates the velocity \(v\) to the distance \(h\) it falls from rest.
Part 2: Falling a Total Distance 'H' to Reach Velocity '2v'
Now, we want to find the total distance the body has to fall from rest (\(u_2 = 0\)) to reach a velocity \(v_{f2} = 2v\). Let this total distance be \(s_2 = H\).
Using the same kinematic equation \(v_{f}^2 = 2gs\):
\[(2v)^2 = 2gH\]
\[4v^2 = 2gH \quad \cdots (2)\]
Relating the Distances
We have two equations:
- \(v^2 = 2gh\)
- \(4v^2 = 2gH\)
Substitute the expression for \(v^2\) from equation (1) into equation (2):
\[4(2gh) = 2gH\]
\[8gh = 2gH\]
Assuming \(g \neq 0\), we can divide both sides by \(2g\):
\[\frac{8gh}{2g} = \frac{2gH}{2g}\]
\[4h = H\]
This means the total distance the body must fall from rest to reach a velocity of \(2v\) is \(4h\).
Finding the Additional Distance
The question asks for the distance it has to fall further. The body has already fallen a distance of \(h\).
Additional distance = Total distance fallen to reach \(2v\) - Distance already fallen to reach \(v\)
Additional distance = \(H - h\)
Substitute the value of \(H = 4h\):
Additional distance = \(4h - h\)
Additional distance = \(3h\)
Summary of Results
A body falling freely from rest acquires velocity \(v\) after falling a distance \(h\), related by \(v^2 = 2gh\).
To acquire a velocity of \(2v\), the body must fall a total distance of \(H\), related by \((2v)^2 = 2gH\), which simplifies to \(4v^2 = 2gH\). Substituting \(v^2 = 2gh\), we get \(4(2gh) = 2gH\), leading to \(H = 4h\).
Since the body has already fallen \(h\), the additional distance required to reach \(2v\) is \(H - h = 4h - h = 3h\).
| Parameter | Initial State (at rest) | After falling 'h' | After falling total 'H' (to reach 2v) |
|---|---|---|---|
| Initial Velocity | \(u=0\) | N/A (this is the final state for the first phase) | \(u=0\) |
| Final Velocity | N/A | \(v_{f1}=v\) | \(v_{f2}=2v\) |
| Distance Fallen (from rest) | N/A | \(s_1=h\) | \(s_2=H\) |
| Kinematic Relation | N/A | \(v^2 = 2gh\) | \((2v)^2 = 2gH\) |
| Derived Total Distance \(H\) | N/A | N/A | \(H=4h\) |
| Additional Distance | N/A | N/A | \(H-h = 4h-h = 3h\) |
Revision Table: Key Concepts for Freely Falling Bodies
| Concept | Description | Relevant Equation (from rest, downward positive) |
|---|---|---|
| Free Fall | Motion under gravity only; constant acceleration \(g\). | \(a = g\) |
| Starting from Rest | Initial velocity is zero. | \(u=0\) |
| Velocity after distance 's' | Velocity achieved after falling distance 's' from rest. | \(v^2 = 2gs\) |
| Distance after time 't' | Distance fallen after time 't' from rest. | \(s = \frac{1}{2}gt^2\) |
| Velocity after time 't' | Velocity achieved after time 't' from rest. | \(v = gt\) |
Additional Information: Energy Conservation in Free Fall
Alternatively, this problem can be solved using the principle of conservation of mechanical energy. For a body falling freely, the mechanical energy (sum of kinetic and potential energy) is conserved, assuming no air resistance.
Let's take the starting point (at rest) as the reference level for potential energy, so potential energy there is zero. Downward motion means potential energy becomes negative.
- Initially, at rest at height 0: Kinetic Energy \(KE_1 = \frac{1}{2}mu^2 = 0\), Potential Energy \(PE_1 = 0\). Total Energy \(E_1 = 0\).
- After falling distance \(h\), velocity is \(v\): Height is \(-h\). \(KE_2 = \frac{1}{2}mv^2\), \(PE_2 = mg(-h) = -mgh\). Total Energy \(E_2 = \frac{1}{2}mv^2 - mgh\).
- After falling total distance \(H\) to reach \(2v\): Height is \(-H\). \(KE_3 = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2)\), \(PE_3 = mg(-H) = -mgH\). Total Energy \(E_3 = \frac{1}{2}m(4v^2) - mgH\).
By conservation of energy, \(E_1 = E_2 = E_3\).
From \(E_1 = E_2\):
\[0 = \frac{1}{2}mv^2 - mgh\]
\[mgh = \frac{1}{2}mv^2\]
\[2gh = v^2 \quad \text{(Same as equation (1))}\]
From \(E_1 = E_3\):
\[0 = \frac{1}{2}m(4v^2) - mgH\]
\[mgH = \frac{1}{2}m(4v^2)\]
\[gH = 2v^2\]
\[H = \frac{2v^2}{g}\]
Substitute \(v^2 = 2gh\) into the equation for \(H\):
\[H = \frac{2(2gh)}{g}\]
\[H = 4h \quad \text{(Same total distance)}\]
The additional distance is still \(H - h = 4h - h = 3h\).
Both methods, using kinematic equations or energy conservation, yield the same result for the additional distance a freely falling body must fall to double its velocity from rest.
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