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# Free Fall Formula

## What is Free Fall Formula?

Last updated date: 31st Jan 2023
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A free-falling object formula describes the self-governing phenomena of the body having some mass. A free-fall concept that talks about the body freely falling under gravity.

Assume that a body with velocity v is descending freely from a mountain of height (h) for time (t) seconds. Now, because of gravity (g), the body falls in the following manner:

From this example, we can describe the free-fall motion formula. As it covers a certain distance, we can describe the free fall distance formula. An object possessing a free-fall object formula bears velocity, which we can calculate by using the free-fall velocity formula.

Since the object falling from a mountain has a maximum height, so does the object. The eight can be calculated by using the maximum height formula free fall. As there is a rate of change of velocity in an object while free-falling, so we can determine the free-fall acceleration formula as well.

So, on this page, we will cover all the equations for a falling body along with the free fall physics formula, and then derive the free fall formula as well.

### Free Fall Physics Formula

We know that any object that is moving and being acted upon only by the force of gravity is said to be "in a state of free fall." Such an object experiences a downward acceleration of 9.8 ms-2

We must note that whether the object is falling or rising towards its peak if it is under the sole influence of gravity, its acceleration value will always remain 9.8 ms-2

So, the free-fall acceleration formula says that ‘a’ always equals ‘g’ under free fall.

### Free Falling Bodies Formula

The free-fall motion formula covers the following equations for a falling body:

• Maximum height formula free fall

• Free-fall velocity formula

The maximum height formula free fall is:

h    =   1/2 gt2

The free velocity formula is:

v2  =  2gh, and

v   =  gt

Now, let us derive free fall formula:

### Free Falling Object Formula

Below are the following kinematic equations for deriving the free-fall motion formula:

First equation: if   =  vi  +  at  ………..(1)

The second equation: d  =  vi + vf/2 . t…..(2)

The third equation: vf2   =  vi2  +  2* a* d…..(3)

The fourth equation: d  =  vit +  1/2 at2…..(4)

Here, we replaced ‘s’ with ‘d’, and ‘d’ is the displacement

vi  and vf are the initial and the final velocities of a falling object

a = acceleration, and

t = time in seconds

### Free Fall Velocity Formula

We must note that the initial velocity of the object will become zero, so the first equation becomes:

vf   =   at

Also, according to the free-fall object formula, ‘a = g,’ so the equation (1) becomes:

vf   =   gt

This free-falling bodies formula is the free-fall velocity formula.

Also, from equation (3), we have:

vf2   =  vi2 +  2* a* d

Or,

vf2   =  2gh…….(5)

This is again the free-fall velocity formula.

### Free Fall Distance Formula

From equation (4), we see that displacement is the height traveled by the falling object. So substituting ‘with ‘h,’  and ‘a’ with ‘a,’ we have:

h  =  vit +  1/2 at2

Putting vi   = 0:

h  =  1/2 gt2

This is the required  free fall equation with height ‘h.’

### Maximum Height Formula Free Fall

From equation (5), we have:

vf2   =  2gh

The maximum height formula free fall is:

h   =   vf2/2g

### Concepts to Free Falling Objects Problem Solving

There are a few concepts of free-fall motion that hold paramount importance when using the equations to analyze free-fall motion. These concepts are as follows:

• A freely falling object experiences an acceleration of 9.8 ms-2.  (Here, the negative sign indicates a downward acceleration or deceleration).Whether clearly stated or not, the value of the acceleration in the kinematic equations remains 9.8 ms-2 for any freely falling object.

• If an object is mistakenly dropped (as opposed to being thrown) from an elevated height, its initial velocity remains 0 m/s.

• If an object is projected upwards in an exactly vertical direction, it slows down as it rises upward. The point at which it reaches the peak of its trajectory is the point where the velocity is 0 m/s. This value can be used as one of the important motion parameters in the kinematic equations; for instance, the final velocity (vf) after traveling to the peak reaches a value of 0 m/s.

• If an object is projected upwards in an exactly vertical direction, its velocity at which it is projected equals in magnitude but in a sign opposite to the velocity after it returns to the same height.

In the nutshell, a ball projected with an upward velocity of + 50 m/s will have a downward velocity of - 50 m/s when it returns to the same height.

Now, let’s apply these concepts in solving problems on a free fall formula:

### Free Fall Formula Calculator

For understanding the free fall formula; let’s have a look at the below examples to apply the equation for the freely falling body:

Example 1: What will be the height of the body if it has a mass of 3 kg and after 8 seconds it reaches the ground?

Solution:

Given data:

Height h =?

Time t = 8s

You are acquainted with the concept that free fall is independent of mass. So, using the free-fall formula here:

h  =  1/2 gt2

Putting g  = 9.8 ms-2 and t = 8s:

h = 1/2 *  9.8* (8)2

On solving, we get:

h = 313.6 m

Answer: Therefore, the maximum height that a body covers to reach the ground is 313.6 m.

Example 2:  The cotton ball falls after 4 s and the iron ball falls after 7 s. Determine which object falls with a higher velocity?

Solution:

Since the velocity in free fall is independent of mass, so apply the following formula:

v (Velocity of cotton ball) = gt = 9.8 m/s2 × 4 s = 39.2 m/s

v (Velocity of iron ball) = gt = 9.8 m/s2 × 7 s = 68.6 m/s

We see that the iron ball falls with a higher velocity than the cotton ball.

### Study Habits Students should adopt for Better Scores

Inculcating good habits in daily routine helps an individual to make their days more productive in many ways. Science is a scoring subject and one can secure a good percentage with required hard work. Many students find science a tough subject but it can surely help you to gain more marks as it is based on logistics and reason. Read on to know a few study habits students can adopt to gain more than 95% in science:

• Maintain different notebooks for your formulas, equations, and theories as it will help students to find what they are looking for easily from the vast sea of the syllabus. This will also help in making the revision process quick and easy.

• While preparing for the science exam make sure to solve all the numericals by yourself first. Surely students can take help from their teachers or books if they are stuck with some question but the first efforts should be made to do it all by yourself and try again and again. Learning and understanding concepts are important but it is also important that students are aware of when these are to be implemented and how.

• Students should analyze while studying the topics they are finding difficult to understand and focus on them. If students are aware of their weak points only then they can focus more on them and make it their strength.

• Regularity is an important habit students must inculcate to secure good marks. A student regular with his/ her syllabus will help them to understand their grasp on concepts. Regularity not only includes attending classes, lectures, and tuitions but also self-study sessions.

• Students should first concentrate on the course books that are provided by the NCERT. Many students have the habit of jumping to reference books before finishing the NCERT. It is an observation that most of the questions asked in board exams are asked from NCERT books. Reference books can be used as a source of extra material or for in-depth study of a particular concept. However, NCERT is a student’s bible during the exam season.

• Students should maintain a notebook for notes. Notes work as a summary of each concept during the revision period. It speeds up the process of revision and is a convenient way to learn from. Students can also include insights from their teachers or their insights for particular topics.

### Conclusion

We call the free fall bodies formula the kinematic equations free fall because they are derived from kinematic equations.

Also, free fall is independent of the mass and it only depends on the height the object fell from.

## FAQs on Free Fall Formula

1. How do you calculate the time of a Falling Object?

Firstly, measure the distance the object covered while freely falling (in feet) with a ruler or measuring tape.

Secondly, divide the measured distance by 16.

For instance, if the object falls from a 128 feet mountain, divide 128 by 16, we get 8.

Now, calculating the square root of the obtained number, i.e.8 results, we get the time it takes the object to fall in seconds.

2. What is the Law of Free Fall?

Galileo's Law of Free Fall Statement: In the absence of air resistance (air friction), all bodies (irrespective of their mass) fall with the same acceleration. We must note that this law is an approximation shown by using Newtonian mechanics.

3. Write three examples of Free Fall.

Close to the surface of the Earth, an object that falls freely in a vacuum accelerates at 9.8 m/s2. If there is an air resistance acting on an object that has been dropped, the object eventually reaches a terminal velocity, which is around 53 m/s (190 km/h or 118 mph) for a human skydiver.

4. Is Class 11 physics difficult?

Subjects are not difficult if one pays attention to the concepts and has the right guidance. With hard work, smart work and continuous practice students will get accustomed to the concepts of Physics. Students may find physics in Class 11 slightly tricky as it is more in-depth with the concepts. There is also a new topic that is covered in the syllabus which may consume a bit more of students’ time. But as students will move forward in chapters, they will be able to connect the concepts and understand the chronology. Students can always refer to NCERT solutions available online to understand the concepts quickly, easily, and in simple language.

5. Which is the easiest chapter of the Class 11 Physics NCERT textbook?

Beginning physics in class 11 can be a bit tricky for students to learn and understand but it is not an impossible job. If a student has the required focus and the will to understand then he/she will find that most of the topics in the syllabus are easy. Students must work hard on understanding the fundamental concepts of all topics. This will help them to learn the topics quickly. The concept of an easy chapter and a difficult chapter is very subjective and depends on the concept of grasping the power of students. What is considered difficult by one student might be an easy concept for another. Therefore, students are advised to focus on understanding the concepts.

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