Calculating speed given acceleration and distance

[Zugag]

Hello all! I am new to the forum, so sorry if I post this into a wrong place.

 

So I have been wondering about what if you would drop a coin, for instance off a building that is 400 meters tall, what would be its velocity when it hits the ground given the acceleration of 9.6 m/s^2 due to gravity. I have seen an other topic about the same problem, but it didn't help me at all. Can somebody please help me?

 

Thank you!

[ajb]

This sounds like a homework question. Thus I won't give you the full answer.

 

What you need to do is look at the equations you know for constant acceleration and see which one will work in this case. That is what do you know and what do you want to calculate. From the question you know the initial velocity of the coin (u=0), the distance travelled (s = 400 m ) and you know a = 9.6 m/s^2... So what about using v^2 = u^2 + 2 a s?

[Zugag]

Thanks a lot for the reply, but it's not a homework question, I am just learning basics for my upcoming course. So did you mean that simplified it is V^2 = 0 + 2×(9.6)×(400)? So for an answer i got V^2= 7680 m/s, and i know that it's completely wrong, so i'm doing something wrong here.

[studiot]

You would need to be about 70 kilometres up in the sky for the acceleration due to gravity to be 9.6 m per second²

 

The standard value is 9.81 ms^-2

 

Often you can use 10 as a good enough approximation.

 

Using this in the formula provided by ajb

 

final speed = sqrt (2 x 10 x 400) m/s = sqrt(8000) = 89.5 m/sc

 

When you do science it is vitally important that you get you physics facts and your arithmetic correct.

Edited February 27, 2015 by studiot

[ajb]

Thanks a lot for the reply, but it's not a homework question, I am just learning basics for my upcoming course. So did you mean that simplified it is V^2 = 0 + 2×(9.6)×(400)? So for an answer i got V^2= 7680 m/s, and i know that it's completely wrong, so i'm doing something wrong here.

Now take the square root. I make that 87.6 m/s. This sounds okay.

[Zugag]

Thanks a lot guys, now I know the technique. Have a great day!

[ajb]

Thanks a lot guys, now I know the technique. Have a great day!

No problem. I am glad I have helped.

[studiot]

For elementary kinematics (involving only constant acceleration) there are not many formulae

 

v² = u² + 2as

 

v = u + at

 

s= ut+ 1/2 (at²)

 

a = (v-u)/t

 

s = 1/2(v+u)t

 

Do you know these?

[Sensei]

So I have been wondering about what if you would drop a coin, for instance off a building that is 400 meters tall, what would be its velocity when it hits the ground given the acceleration of 9.6 m/s^2 due to gravity. I have seen an other topic about the same problem, but it didn't help me at all. Can somebody please help me?

Look at this video on YouTube

This is exactly about subject you're talking about showed in OpenOffice SpreadSheet. Which you can download for free from http://www.openoffice.org