A state in which a body moves solely under the influence of the earth’s gravity is known as fall. The object’s motion will be accelerated as a result of the acceleration generated by this external force on the item. As a result, fall motion is also known as gravitational acceleration.

When dropped from a certain height in the absence of air, a coin and a feather reach the ground simultaneously. A coin is heavier than a feather, so how does both take the same time to reach the ground? It is because both of them are under-FALL. With examples in this article, we will learn about the Fall Objects, Motion Under Gravity Fall.

What is fall?

Freefall is defined as the condition in which the only force acting on a body is due to gravity, and hence its acceleration is due to gravity, which is \(g = 9.8\;{\rm{m}}{{\rm{s}}^{ – 2}}\). Since the value of \(g\) is constant, motion under fall is an example of uniformly accelerated motion. fall motion can be analysed using the equations of motion for constant acceleration.

A unique but exciting property of the acceleration due to gravity is that it is the same for all masses. This was far from the self-evident fact until the days of Galileo Galilei. Galileo conducted these ingenious experiments at the “leaning” Tower of Pisa and proved by dropping masses of different weights from the top of the tower that gravitational acceleration is independent of the mass of the objects.

In physics, problems onfall are based on the assumption of the absence of air resistance. However, the Earth’s atmosphere provides some resistance to an object in fall in the real world. Also, particles in the air collide with the falling object, transforming their kinetic energy into thermal energy. It results in slowly increasing downward velocity.

Equations of Motion (Freefall)

The equations of motion of an objectly falling under gravity can be obtained by substituting the acceleration value equal to the acceleration due to gravity. Consider an object fallingly from a height \(h\), travels for time \(t\) seconds, acquiring a final velocity \(v\), under the acceleration due to gravity \(g\). Since the object’s initial velocity is zero, substitute \(u = 0\) in the equations to motions, and we get:
1. \(v = gt\)
2. \(h = \frac{1}{2}g{t^2}\)
3. \({v^2} = 2gh\)

Calculation of Acceleration Due to Gravity

A body falling only under the Earth’s gravitational force is aly falling body. Ideally, when an object is thrown from a height, it encounters air resistance forces along with the gravitational force. For simplicity, we assume no force due to air resistance acting on the body.  Whenever an object falls, there is no change in the object’s direction of motion. However, there will be a change in velocity, and hence there will be acceleration. This acceleration due to Earth’s gravitation is called acceleration due to gravity and is represented by \(g\).

The gravitational force on an object kept at a height \(h\) above the surface of the Earth is equal to \(F = \frac{{GMm}}{{{{(R + h)}^2}}}\)

Where \(G\) is the universal gravitation constant, \(M\) is the mass of the Earth, \(m\) is the object’s mass, and \(R\) is the Earth’s radius.

If \(h < < R,\,F = \frac{{GMm}}{{{{(R)}^2}}}\)

The weight of the object will balance the gravitational force acting on the object, \(W = m \cdot g\).

This means, \(W = F\)

\(mg = \frac{{GMm}}{{{{(R)}^2}}}\)
\(g = \frac{{GM}}{{{{(R)}^2}}}\)

For an object falling solely under the Earth’s gravitational force, the object’s acceleration does not depend on the mass of an object. Heavier as well as light objects will have the same acceleration.

1. The Earth’s radius at the poles is minimum, the value of \(g\) is maximum at poles.
2. The radius of Earth at the equator is maximum, the value of \(g\) is minimum at the equator.
3. The value of \(g\) is maximum at the Earth’s surface, and it decreases as we go up towards the higher altitudes.
4. Although the value of \(‘g’\) is not constant, it varies with the object’s position on Earth’s surface. The generally accepted value of g is \(9.8\;{\rm{m}}/{{\rm{s}}^2}\)
5. The value of \(g\) on the moon is almost \(\frac{1}{6}\) times the value of \(g\) at the Earth.

Problems Based on Fall

Q.1. Compute the body’s height if it has a mass of \({\rm{5}}\,{\rm{kg}}\) and touches the ground after \(10\) seconds?
Ans: Given parameters are:
Time \(t = 10\) sec
We have to compute the height. So, we can apply the first equation as given above.
i.e., \(h = \frac{1}{2}g{t^2}\)
Substituting the values,
\(h = \frac{1}{2} \times 9.8 \times {(10)^2}\)
\(h = 4.9 \times {(10)^2}\)
\(h = 490\;{\rm{m}}\)

Q.2. A ball is thrown vertically upwards with a velocity of \(49\;{\rm{m}}/{\rm{s}}\). Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the Earth.
Ans:
(i) Initial velocity, \(u = 49\;{\rm{m}}/{\rm{s}}\)
Final velocity, \(v = 0\)
\(a = g =\,- 9.8\;{\rm{m}}/{{\rm{s}}^2}\)
Height \(=\) Distance \( = s = \) ?
\({v^2} – {u^2} = 2gs\)
\(0\,- {(49)^2} = 2 \times – 9.8 \times s\)
\(s = \frac{{ – 49 \times 49}}{{2 \times – 9.8}} = 122.5\;{\rm{m}}\)
(ii) Time take \(t =\)?
\(v = u + gt\)
\(\therefore \,0 = 49 + ( – 9.8) \times t\)
\(t = \frac{{ 49}}{{9.8}} = 5\,\rm{s}\)
The total time taken to return the Earth’s surface by the ball is \(5\,\rm{s} + 5\,\rm{s} = 10\,\rm{s}\).

Q.3. A stone is released from the top of a tower of height \(19.6\;{\rm{m}}\). Calculate its final velocity just before touching the ground?
Ans: According to the question,
\(u = 0\;{\rm{m}}/{\rm{s}}\)
\(v = \)?
\(h = s = 19.6\;{\rm{m}}\)
\(g = 9.8\;{\rm{m}}/{{\rm{s}}^2}\) (falling down)
\({v^2} – {u^2} = 2gs\)
\({v^2} – {(0)^2} = 2 \times 9.8 \times 19.6\)
\(v = 19.6\;{\rm{m}}/{\rm{s}}\)
The final velocity just before touching the ground is \(19.6\;{\rm{m}}/{\rm{s}}\).

Summary

Freefall is defined as the condition in which the only force acting on a body is due to gravity, and hence is denoted as \(g = 9.8\,{\rm{m}}{{\rm{s}}^{{\rm{ – 2}}}}\).  Freefall physics problems are based on an assumption of the absence of air resistance. However, the Earth’s atmosphere provides some resistance to an object in fall in the real world.

Whenever an object falls, there is no change in the object’s direction of motion. However, there will be a change in velocity, and hence there will be acceleration. This acceleration due to Earth’s gravitation is called acceleration due to gravity and is represented by ‘\(g\)’. It can be given as: \(g = \frac{{GM}}{{{{(R)}^2}}}\)

For an object falling solely under the Earth’s gravitational force, the object’s acceleration does not depend on the mass of an object. Heavier as well as light objects will have the same acceleration.  From the formula, we can see that acceleration due to gravity varies with its distance from Earth’s surface.

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FAQs

Q1. What do you mean by fall?
Ans. Whenever an object falls toward Earth under the force of gravity only, and no other force is present, the object’s motion is said to be “fall.”

Q2. Why is the weight of an object on the moon 1/6th its weight on the Earth?
Ans. The weight of an object depends on ‘\(g\)’ acceleration due to gravity, and the value of ‘\(g\)’ on Earth and the moon are not the same.
The acceleration due to gravity on the moon is one-sixth the acceleration due to gravity at Earth. That is why the weight of a body on the Earth is six times more than the weight of the same body on the moon.

Q3. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans. A sheet of paper has a larger surface area, but its surface area decreases when crumpled into a ball. While falling, it has to overcome the force exerted by air/wind called air resistance. The force due to air resistance increases as the surface area of an object increases. Hence the sheet of paper falls slower. 

Q4. What do you mean by the acceleration due to gravity?
Ans. The acceleration of fall is the acceleration due to gravity. We can also say the acceleration of an object due to Earth’s gravitational force acting on it is known as acceleration due to gravity. 

Q5. Gravitational force acts on all objects in proportion to their masses. Why, then, a heavy object does not fall faster than a light object?
Ans. The acceleration due to gravity ‘\(g\)’ acting on a heavy object is independent of the body’s mass.
\(g = \frac{{GM}}{{{{(R)}^2}}}\) and \(F = \frac{{GMm}}{{{{(R)}^2}}}\)
The gravitational force acting on an object depends on its mass.

Q6. What is the acceleration of fall?
Ans. When an object falls due to Earth’s gravitational pull, its velocity changes, and It accelerates due to the Earth’s gravity; it is said to be under the acceleration of fall. This acceleration is calculated to be \(9.8\;{\rm{m}}/{{\rm{s}}^2}\).

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