# Question on velocity changes in free fall

• Physiophysio
In summary: The ball still has the same mass as before, so its weight is still mg. The only difference between the two situations is that now there is an additional force directed upwards, equal to mg/2. As a result, since the net force is now F = mg - mg/2 = mg/2, the acceleration is only half as large as in the previous situation.
Physiophysio

## Homework Statement

Assume a ping pong of mass 3g is falling, and it has fallen 1m and is now traveling at 5 m/s. If the air resistance force is half its weight, how far will it travel in the next 0.1 seconds. [Hint: you may want to sketch a velocity-time graph)

## Homework Equations

Understanding that area under v-t graph is distance travelled.
F=ma
a=v/t
v=d/t

## The Attempt at a Solution

So I thought:
If mass=3g, then weight is 0.03N
And as air resistance is half that, then the resultant force on the ping pong ball is 0.03-(0.03/2)=0.015N
Therefore, as F=ma,
0.015=m * a
But we know m=0.003 kg, so,
0.015=0.003 * a
a=5 metres per second per second.

Hence, in 0.1 seconds,

5 * 0.1 = 0.5 m/s

So the velocity of the ping pong ball increases by 0.5 m/s in the next 0.1 seconds.
And we know initial velocity was 5 m/s, so it increased to 5.5 m/s in 0.1 seconds.

So on a velocity-time graph, this forms a trapezium shape, and you find the area of that trapezium to find the distance the ping pong ball has traveled in 0.1 seconds:

[ (5+5.5)/2 ] * 0.1 = 0.525 m.

So the ping pong ball travels 525 centimetres (or 0.525 metres) in the next 0.1 seconds.

Is that right?

Thanks :)

Physiophysio said:

## Homework Statement

Assume a ping pong of mass 3g is falling, and it has fallen 1m and is now traveling at 5 m/s. If the air resistance force is half its weight, how far will it travel in the next 0.1 seconds. [Hint: you may want to sketch a velocity-time graph)

## Homework Equations

Understanding that area under v-t graph is distance travelled.
F=ma
a=v/t
v=d/t

## The Attempt at a Solution

So I thought:
If mass=3g, then weight is 0.03N
And as air resistance is half that, then the resultant force on the ping pong ball is 0.03-(0.03/2)=0.015N
Therefore, as F=ma,
0.015=m * a
But we know m=0.003 kg, so,
0.015=0.003 * a
a=5 metres per second per second.

Hence, in 0.1 seconds,

5 * 0.1 = 0.5 m/s

So the velocity of the ping pong ball increases by 0.5 m/s in the next 0.1 seconds.
And we know initial velocity was 5 m/s, so it increased to 5.5 m/s in 0.1 seconds.

So on a velocity-time graph, this forms a trapezium shape, and you find the area of that trapezium to find the distance the ping pong ball has traveled in 0.1 seconds:

[ (5+5.5)/2 ] * 0.1 = 0.525 m.

So the ping pong ball travels 525 centimetres (or 0.525 metres) in the next 0.1 seconds.

Is that right?

Thanks :)
Hello Physiophysio. Welcome to PF !

That looks like it's as good an approximation as you can get with the given information.

Very good analysis.

Physiophysio
Yes. (You don't actually need the mass. You know the acceleration will be g/2.)

SammyS said:
Hello Physiophysio. Welcome to PF !

That looks like it's as good an approximation as you can get with the given information.

Very good analysis.
Thanks :)

haruspex said:
Yes. (You don't actually need the mass. You know the acceleration will be g/2.)
Could you please explain why that is, because, yes, I see how a=5 which is g/2, but how does that come about? Is it because, in free fall with no air resistance and only the weight, a=g, but now air resistance is half of weight, so a would be a/2 which is equal to g/2?
Or is that logic wrong? Thanks for replying!

Physiophysio said:
so a would be a/2 which is equal to g/2?
If a is defined as the acceleration it undergoes, a is still a, but is now g/2, yes.

Physiophysio

## 1. How does velocity change in free fall?

In free fall, an object experiences a constant acceleration of 9.8 m/s^2 due to the force of gravity. This means that its velocity increases by 9.8 m/s every second. However, this acceleration only occurs in the vertical direction, not in the horizontal direction.

## 2. Does the mass of an object affect its velocity in free fall?

No, the mass of an object does not affect its velocity in free fall. This is because in free fall, all objects experience the same acceleration due to gravity regardless of their mass.

## 3. How does air resistance impact velocity in free fall?

Air resistance can affect the velocity of an object in free fall. As an object falls, it will eventually reach a point where the force of air resistance becomes equal to the force of gravity. At this point, the object will reach its terminal velocity and will no longer accelerate.

## 4. Can an object have a constant velocity in free fall?

No, an object cannot have a constant velocity in free fall. This is because in free fall, an object is constantly accelerating due to the force of gravity. However, once it reaches its terminal velocity, its velocity will remain constant until it reaches the ground.

## 5. How can the velocity of an object in free fall be calculated?

The velocity of an object in free fall can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the object has been falling. This equation assumes that the object starts from rest.

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