Motion of Body Under Gravity (Free Fall)

It is a basic concept in physics, indicating the motion of a body under gravity when the force of gravity acts on that body. This type of motion is commonly referred to as free fall. That is, if the object is dropped from a height, it accelerates towards the Earth on account of gravity, which acts with a constant acceleration of about 9.8 m/s².

This principle applies in a much greater variety of examples—from an apple dropping from a tree to a skydiver diving out of an aeroplane. In each case, the object has a uniform acceleration in the same direction as that due to gravity. The equations of motion enable one to find the time it takes for an object to hit the ground, its velocity at any time, and the distance covered.

In this article, we will cover the motion of the body under gravity. This concept falls under the broader category of kinematics which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of 15 questions have been asked on this concept. And for NEET 4 questions were asked from this concept.

What Is Acceleration Due to Gravity and Sign Convention

The force of attraction of the earth on anybody is called the force of gravity. Acceleration produced on the body by the force of gravity is called acceleration due to gravity. It is represented by the symbol ‘g’.

Upward direction and right direction are taken as positive

The downward direction and left direction is taken as negative

There are Three Cases Basically in This

Case 1: If a body dropped from some height (initial velocity zero)

$\begin{aligned}
& \mathrm{u}=0 \\
& \mathrm{a}=\mathrm{g} \\
& v=g t \\
& h=\frac{1}{2} g t^2 \\
& v^2=2 g h \\
& h_n=\frac{g}{2}(2 n-1)
\end{aligned}$

Case 2: If a body is projected vertically downward with some initial velocity

$ \begin{aligned}
& \text { Equation of motion: } \quad v=u+g t \\
& \begin{array}{ll}
h= & u t+\frac{1}{2} g t^2 \\
v^2 & =u^2+2 g h \\
h_n \quad & =u+\frac{g}{2}(2 n-1)
\end{array}
\end{aligned} $

Case 3: If a body is projected vertically upward.

(i) Apply equation of motion:

Take initial position as origin and the direction of motion (vertically up) as $\mathrm{a}=-\mathrm{g}$ [as acceleration due to gravity is downwards]
So, if the body is projected with velocity $u$, and after time $\mathrm{t}$ it reaches up to height $\mathrm{h}$ then,
$
\mathrm{v}=\mathrm{u}-\mathrm{gt} ; \quad \mathrm{h}=\mathrm{ut}-\frac{1}{2} g \mathrm{t}^2 ; \mathrm{v}^2=u^2-2 \mathrm{gh}
$

(ii) For the case of maximum height $v=0$

So from the above equation

$\begin{gathered}
u=g t \\
h=\frac{1}{2} g t^2 \\
\text { and } u^2=2 g h
\end{gathered}$

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Solved Example Based on Motion of Body Under Gravity

Example 1: A balloon is moving up in the air vertically above point $\mathrm{A}$ on the ground. When it is at a height $h_1$, a girl standing at a distance (point $B$ ) from $A$ (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h_2$, it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point $C$ ). Then the height $h_2$ is (given $\tan 30^{\circ}=0.5774$ ) :

1)1.464d
2) $0.732 d$
3) $0.464 d$
4) $d$

Solution:

$\begin{aligned}
& \frac{\mathrm{h}_1}{\mathrm{~d}}=\tan 45^{\circ} \Rightarrow \mathrm{h}_1=\mathrm{d} \ldots \\
& \frac{h_1+h_2}{d+2.464 d}=\tan 30^{\circ} \\
& \Rightarrow\left(h_1+h_2\right) \times \sqrt{3}=3.46 \mathrm{~d} \\
& \left(\mathrm{~h}_1+\mathrm{h}_2\right)=\frac{3.46 \mathrm{~d}}{\sqrt{3}} \\
& \Rightarrow \mathrm{d}+\mathrm{h}_2=\frac{3.46 \mathrm{~d}}{\sqrt{3}} \\
& \mathrm{~h}_2=\mathrm{d}
\end{aligned}$

Hence, the answer is option (4).

Example 2: When a ball is dropped anto a lake from a height $4.9 \mathrm{~m}$ above the water level, it hits the water with a velocity $v$ and then sinks to the bottom with the constant velocity $v$. It reaches the bottom of the lake 4.0 safter it is dropped. The approximate depth of the lake is :
1) $19.6 \mathrm{~m}$
2) $29.4 \mathrm{~m}$
3) $39.2 \mathrm{~m}$
4) $73.5 \mathrm{~m}$

Solution:

$\mathrm{v= \sqrt{2gh}= \sqrt{2\times 9.8 \times 4.9}= 9.8\, m/s}$
$\mathrm{\text{time for A to B}= \sqrt{\frac{2h}{g}}= \sqrt{\frac{2\times4.9}{9.8}}= 1\, sec}$
distance travelled by ball in 3 sec in lake $\mathrm{= 9.8\times 3}$
$\mathrm{\therefore }$ depth of lake $\mathrm{= 29.4\, m}$

The correct answer is (2)

Example 3: A tennis ball is released from a height $h$ and after freely falling on a wooden floor, it rebounds and reaches height $h$ $\overline{2}$. The velocity versus height of the ball during its motion may be represented graphically by:
(graphs are drawn schematically and not to the scale)

1)

2)

3)

4)

Solution:

Velocity at the ground (means zero height) is non-zero therefore the option 4 is incorrect.

The velocity versus height curve will be parabolic as for motion under gravity.
i.e The velocity versus height curve is non-linear therefore option 1 is also incorrect.
$
\begin{aligned}
& \mathrm{v}^2=2 \mathrm{gh} \\
& \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dh}}=2 \mathrm{~g}=\mathrm{const} \\
& \Rightarrow \frac{\mathrm{dv}}{\mathrm{dh}}=\frac{\text { constant }}{\mathrm{v}}
\end{aligned}
$

Here we can see the slope is very high when velocity is low therefore at the maximum height the slope should be very large which is in option 3 and as velocity increases slope must decrease therefore option 3 is correct.

Example 4: An NCC parade is going at a uniform speed of $9 \mathrm{~km} / \mathrm{h}$ under a mango tree on which a monkey is sitting at a height of $19.6 \mathrm{~m}$. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ )

1)$5 \mathrm{~m}$

2)$10 \mathrm{~m}$

3)$19.8 \mathrm{~m}$

4)$24.5 \mathrm{~m}$

Solution:

$
\begin{aligned}
& \text { Time to fall }=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=2 \mathrm{~s} \\
& \begin{aligned}
& \text { Velocity at parade }=9 \frac{\mathrm{km}}{\mathrm{hr}} \\
&=9 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\
&=2.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
$

Cadet will receive mango at a distance $\mathrm{x}$
$
\mathrm{x}=\mathrm{v}_{\mathrm{x}} \mathrm{t}=2.5 \times 2=5 \mathrm{~m}
$

Hence 1 is correct option.

Example 5: Two balls $\mathrm{A}$ and $\mathrm{B}$ are placed at the top of $180 \mathrm{~m}$ tall tower. Ball $A$ is released from the top at $t=0 \mathrm{~s}$. Ball $\mathrm{B}$ is thrown vertically down with an initial velocity' $\mathrm{u}^{\prime}$ at $\mathrm{t}=2 \mathrm{~s}$. After a certain time, both balls meet $100 \mathrm{~m}$ above the ground. Find the value of ' $\mathrm{u}$ ' in $\mathrm{ms}^{-1}$. [use $g=10 \mathrm{~ms}^{-2}$ ]:

1)$10$

2)$15$

3)$20$

4) $30$

Solution:

Let they meet $t=t_{\circ}$
For A
$
80=\frac{1}{2} g t_0^2
$

For B
$
80=u\left(t_0-2\right)+\frac{1}{2} g\left(t_0-2\right)^2
$

From (i)
$
\mathrm{t}_0=4
$
$
\begin{aligned}
& 80=\mathrm{u} \times 2+5 \times 2^2 \\
& \mathrm{u}=30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

The correct answer is Option (4)

Summary

Understanding the motion of a body under gravity means being able to explain very many daily phenomena, such as why objects fall when dropped and how projectiles travel.

It is a universal theory that has broad use, ranging from the prediction of a path a basketball will take when it is thrown to explaining how satellites orbit. Consequently, all objects should respond equally to the force of gravity, thus having uniform acceleration while in motion, which makes the motion of the falling objects predictable and consistent.