Zugag Posted February 27, 2015 Share Posted February 27, 2015 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! Link to comment Share on other sites More sharing options...
ajb Posted February 27, 2015 Share Posted February 27, 2015 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? Link to comment Share on other sites More sharing options...
Zugag Posted February 27, 2015 Author Share Posted February 27, 2015 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. Link to comment Share on other sites More sharing options...
studiot Posted February 27, 2015 Share Posted February 27, 2015 (edited) 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 Link to comment Share on other sites More sharing options...
ajb Posted February 27, 2015 Share Posted February 27, 2015 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. Link to comment Share on other sites More sharing options...
Zugag Posted February 27, 2015 Author Share Posted February 27, 2015 Thanks a lot guys, now I know the technique. Have a great day! Link to comment Share on other sites More sharing options...
ajb Posted February 27, 2015 Share Posted February 27, 2015 Thanks a lot guys, now I know the technique. Have a great day! No problem. I am glad I have helped. Link to comment Share on other sites More sharing options...
studiot Posted February 27, 2015 Share Posted February 27, 2015 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? 1 Link to comment Share on other sites More sharing options...
Sensei Posted February 28, 2015 Share Posted February 28, 2015 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 Link to comment Share on other sites More sharing options...
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