Ball dropped in air at 20m/s, why v=20 not v0=20

[MrGoATi]

Homework Statement

ball dropped in air at 20m/s.
why is v=20,v₀=0 instead of v₀=20,v=0

Homework Equations

a=v-v₀ / t
s=x₀+v_0xt + at²/2

The Attempt at a Solution

a=g=9.8m/s²

I know I can solve it with first finding time(I get that) a=v-v₀/t making it into t=v-v₀/g
so 20/9.8=2.04s
but now from what I understand 20m/s is v₀ is it not? can someone explain is it or is it not initial velocity, and why we don't use
s=v_0xt+ gt²/2
but instead s=gt²/2

I know the answer is 20.4m, or maybe it's actually wrong?

 

[Chestermiller]

Can you please supply the exact problem statement.

 

[MrGoATi]

Can you please supply the exact problem statement.

the task is to find maximum height,

but I'm pretty sure it's 20.4 I am asking about whether velocity is v₀ or v . and how do I know which is it in tasks I may have in future.

 

[TomHart]

v_o typically means initial velocity, and v (or v_f) typically means final velocity. But what is not clear to me is what is happening in this problem. You say that the ball is "dropped in air at 20 m/s". That statement itself is confusing. Add to it that you are trying to find maximum height. Is it the case that the ball is being thrown upward from the ground at 20 m/s? If that is true, then the initial velocity (v_o) would be 20 m/s, and the final velocity (velocity at maximum height, that is) would be 0 m/s.

Welcome to Physics Forums.

 

[MrGoATi]

yeah exactly but from a ton of forums I watched solutions to this they use
s=v_0xt+ gt²/2
but end up with s=gt²/2
finding that a ball thrown up with 20m/s will get to ground to maximum height in 2.04s, 20.4m
the task is quite easy, but since it is so confusing for me that v₀=0 I think I will be just as confused in exams when I see anything else than this with velocities and acceleration.

 

[TomHart]

yeah exactly but from a ton of forums I watched solutions to this they use
s=v0xt+ gt2/2
but end up with s=gt2/2

If the equation reduces to s=gt²/2, that would be for a problem where the initial velocity was 0. That is why the v_ot term went away - as would be the case if a ball was dropped off of a building - dropped meaning that it started with an initial velocity (v_o) of 0 m/s.

 

[PeroK]

yeah exactly but from a ton of forums I watched solutions to this they use
s=v_0xt+ gt²/2
but end up with s=gt²/2
finding that a ball thrown up with 20m/s will get to ground to maximum height in 2.04s, 20.4m
the task is quite easy, but since it is so confusing for me that v₀=0 I think I will be just as confused in exams when I see anything else than this with velocities and acceleration.

Suppose a ball is thrown upwards at initial speed $v _0$ and reaches its maximum height, $s$ at $t_1$ seconds. Then:

$t_1 = \frac{v_0}{g}$, hence $v_0 = gt_1$

And:

$s = v_0 t_1 - \frac{1}{2}{g}t_1^2$

Now, substituting $v_0 = gt_1$ into this equation gives:

$s = gt_1^2 - \frac{1}{2}{g}t_1^2 = \frac{1}{2}{g}t_1^2$

 

[MrGoATi]

thanks, now i get it. explains a lot

 

[TomHart]

thanks, now i get it. explains a lot

Me too.

 

[Chestermiller]

MrGoATi: If you had provided the exact problem statement like I asked, this confusion could have been avoided. Please, in the future, provide the complete problem statement, and not your own interpretation.

 

[MrGoATi]

MrGoATi: If you had provided the exact problem statement like I asked, this confusion could have been avoided. Please, in the future, provide the complete problem statement, and not your own interpretation.

i did completely state all that was given, and when you asked i answered what i needed to find. thanks for suggestion though i will keep that in mind

 

[Chestermiller]

i did completely state all that was given, and when you asked i answered what i needed to find.

What you stated was not the exact statement of the problem as it was originally written. It was your paraphrasing of the problem statement.

 

[MrGoATi]

[ post: 5642023, member: 345636"]What you stated was not the exact statement of the problem as it was originally written. It was your paraphrasing of the problem statement.[/QUOTE]
,,ball dropped in air at 20m/s.,, that's the given information the question was ,maximum height, which i told when you asked. i could not tell you exact statement since it is written in Lithuanian. but there's no more info. it's nine word task.

 

[Chestermiller]

[ post: 5642023, member: 345636"]What you stated was not the exact statement of the problem as it was originally written. It was your paraphrasing of the problem statement.

,,ball dropped in air at 20m/s.,, that's the given information the question was ,maximum height, which i told when you asked. i could not tell you exact statement since it is written in Lithuanian. but there's no more info. it's nine word task.[/QUOTE]
Yikes. Whoever stated the problem for you in that way did you a great disservice. They should be berated. I hope this doesn't always happen.

Chet