Question on velocity changes in free fall

[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 :)

 

[SammyS]

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.

 

[haruspex]

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

 

[Physiophysio]

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 :)

 

[Physiophysio]

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!

 

[haruspex]

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.