# Why doesn't my result yield 9.8 m/s^2?

• Henrybar
In summary, a ball falling from rest and located 45 meters below its starting point after 3.0 seconds has an acceleration of 10m/s^2, assuming uniform acceleration and negligible air resistance. This value may differ from the standard gravity value of 9.8m/s^2 used in introductory classes. The relevant equation for this situation is Δd=(v1 x Δt)+(a x Δt^2)/2.
Henrybar
A ball falling from rest is located 45meters below its starting point 3.0s later. Assuming its acceleration is uniform, what is it's value? (air resistance is negligible)

Relevant Equations - Equations for uniformly accelerated motion
Δd=(v1 x Δt)+(a x Δt^2)/2

Attempt To solve:
v1= 0
Δd=45m
Δt=3.0s
a=?

a=(2Δd/Δt^2)-(2v1Δt/Δt^2)
a= ((2 x 45)/9)-((0 x 3)/9)
a=10m/s^2

If a ball is falling from rest(0 initial velocity) shouldn't it's acceleration be 9.8 seconds since gravity is constant? Is my equation wrong? Is my idea of time-acceleration misplaced?

The given values result in the acceleration that you calculated. Your calculation is okay.

In introductory classes g is often approximated to be 10 m/s2 because it simplifies numerical calculations while still getting across the concepts being taught.

The question doesn't say the ball was near the surface of the earth. Don't make unjustified assumptions!

## 1. Why is the acceleration of my object not exactly 9.8 m/s^2?

There are a few possible reasons for this. One reason could be measurement error. It is important to ensure that your measurements are precise and accurate. Another reason could be external factors such as air resistance or friction, which can affect the acceleration of your object. Additionally, the acceleration due to gravity can vary slightly depending on location and altitude on Earth.

## 2. Can the acceleration of an object ever be greater than 9.8 m/s^2?

Yes, the acceleration of an object can exceed 9.8 m/s^2 under certain circumstances. For example, if the object is in free fall, it will experience an acceleration of 9.8 m/s^2 due to gravity. However, if the object is being accelerated by an external force, such as a rocket engine, it can achieve acceleration greater than 9.8 m/s^2.

## 3. Why is 9.8 m/s^2 considered the standard acceleration due to gravity?

The value of 9.8 m/s^2 for the acceleration due to gravity is an approximation and is often used as a standard value for calculations. It is the average value of the acceleration due to gravity at Earth's surface, but it can vary slightly depending on location and altitude. Other planets and celestial bodies have different values for their acceleration due to gravity.

## 4. What factors can affect the acceleration of an object?

The acceleration of an object can be affected by various factors, including the mass of the object, the force acting on the object, and external forces such as air resistance or friction. The acceleration can also be affected by the gravitational pull of other objects, such as the Earth or other celestial bodies.

## 5. Can the acceleration of an object change over time?

Yes, the acceleration of an object can change over time. This is known as changing acceleration and occurs when the object experiences a change in velocity. For example, if a car starts from rest and then speeds up, its acceleration will change over time. This is because its velocity is increasing, and therefore its acceleration is also changing.

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