Video Lesson

Simulations

https://iwant2study.org/lookangejss/02_newtonianmechanics_2kinematics/ejss_model_freefall01/freefall01_Simulation.xhtml

http://sahyun.net/physics/html5/NGPET_freefall/

Lesson Objective

Dear Learners,

By the end of this section, you should be able to:-

• Derive the equations of motion with constant acceleration;
• Use appropriate equations of motion to solve motion-related problems.
• The equations of motion with constant acceleration describe the relationship between an object’s initial velocity (\( u \)), final velocity (\( v \)), acceleration (\( a \)), time taken (\( t \)), and displacement (\( s \)). There are three main equations:

Brainstorming Question

How can you derive the equations of a uniformly accelerated motion?

key terms and concepts

1. Displacement equation:

1. v = u + at
2. S=$\frac{1}{2} \left( u+v \right)$t
3. S=ut+$\frac{1}{2}at^{2}$
4. $V^{2}=U^{2}$+2as
5. S=vt-$\frac{1}{2} at^{2}$

proof equations

(1) First equation of Motion:

V = u + at

solution

Consider a body of mass “m” having initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:

Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

=> a = (v-u) /t
=>at = v-u=>

v = u + at This is the first equation of motion…………………….(1)

(2) Second equation of motion:

sol.
Let the distance travelled by the body be “s”.
We know that

Distance = Average velocity X Time

Also, Average velocity = (u+v)/2

.: Distance (s) = …….eq.(1)

Again we know that:

v = u + at

substituting this value of “v” in eq.(1), we get

This is the 2nd equation of motion……………………………………..(2)

(3) Third equation of Motion

We know that

V = u + at
=> v-u = at
or t = (v-u)/a ………..eq.(2)

Also we know that

Distance = average velocity X Time

This is the third equation of motion………………………………(3)

Exercise 1:

A ball is thrown vertically upward with an initial velocity of 20 m/s. Calculate the maximum height reached by the ball and the time it takes to reach that height. Take acceleration due to gravity as -9.8 m/s$^{2}$.

Solution:

Given:
Initial velocity (u) = 20 m/s
Acceleration due to gravity (a) = -9.8 m/s$^{2}$

To find the maximum height reached (h):
Using the equation v$^{2}$ = u$^{2}$ + 2as, where v = at maximum height,
= (20)$^{2}$ + 2(-9.8)h
-400 = -19.6h
h ≈ 20.41 meters

To find the time taken to reach maximum height:
Using the equation v = u + at, where v = at peak,
= 20 – 9.8t
t ≈ 20 / 9.8
t ≈ 2 seconds

So, the ball reaches a maximum height of approximately 20.41 meters and takes about 2 seconds to reach that height.

Exercise 2:

A rocket accelerates from rest at a rate of \(3 \frac{m}{s^{2}}\) for \(5 \ s\). Calculate its final velocity and the distance it has traveled during this time.

Solution:

Given:
Initial velocity (u) = 0\(\ m/s\) (as rocket starts from rest)
Acceleration (\(a\))= \(3 \frac{m}{s^{2}}\)
Time (\(t\))= \(5 s\)

To find final velocity (\(v\)):

Using the equation:

\[v= u+at,\]
\[v=0+(3)(5),\
v=15 \frac{m}{s}.\]

To find distance traveled (\(s\)):

Using the equation:

$s=ut+\frac{1}{2}at^{2}$
s=(0)(5)+$(\frac {1}{2})(3)(5)^{2}$
s=37.5m

Therefore, after accelerating for \(5 \ s,\)the rocket’s final velocity will be \(15 \frac{m}{s}\), and it will have traveled approximately\(37.5m.\)

Free fall

Free fall refers to the motion of a body where gravity is the only force acting upon it. In such a scenario, the object accelerates downward at a constant rate, which on Earth is approximately (9.8m/s$^{2}$). This acceleration due to gravity is often denoted by ( g ).Key points about free fall include: Acceleration: The acceleration of an object in free fall is constant and equal to ( g ).Velocity: The velocity of the object increases linearly with time, given by ( v = gt ) if it starts from rest. Displacement: The displacement (distance fallen) can be calculated using ( s = $\frac{1}{2}gt^2$ ) when starting from rest. Independence of Mass: In the absence of air resistance, all objects fall at the same rate regardless of their mass.

height ’h’ since the vertical distance of the freely falling bodies is known as
height ’h’.
The equation of motion can thus be modified as:

$v = v_{o} + g t$
$h = v_{o}t +\frac{1}{2}gt ^{2}$
$v^{2}= v^{2}_{0} +2gh$

Example

A mango fruit has fallen from a tree. Find its velocity and vertical height when it reached the ground if it took 1 s to reach the ground.

solution

To find the velocity and vertical height of the mango fruit when it reached the ground, we can use the equations of motion under gravity.

The equation for velocity (\(v\)) in terms of initial velocity (\(v_{o}\)), acceleration due to gravity (\(g\)), and time (\(t\)) is:
\[ v = v_{o} + gt \]

The equation for vertical height (\(h\)) in terms of initial height (\(h_{o}\)), initial velocity, time, and acceleration due to gravity is:
\[ h = h_{o} + v_{o}t + \frac{1}{2}gt^{2} \]

Given that the mango fell from rest (initial velocity \(v_{}=\)), and assuming that it falls near Earth’s surface where \(g = 9.81 \, m/s^2\) is taken as an approximation for acceleration due to gravity, we can calculate:

Velocity:

\[ v = + (9.81)(1) = 9.81 \, m/s \]

Vertical Height:
\[ h = 0 *1 + \frac{1}{2}(9.81)(1)^{2} = 4.905 \, m \]

Therefore, when the mango fruit reaches the ground after falling for 1 second,
– Its velocity will be \(9.81m/s\) downwards
– It will have fallen a total vertical distance of \(4.905m\) from its starting point.

These calculations assume no air resistance affecting the fall.