Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
What am I doing wrong with this physics question?
Here's the question:
The observation deck of a tower is 190 m above the street. Determine the time required for a penny to drop from the deck to the street below. (Answer given: 6.22 s.)
I used (excuse the lack of deltas and vector symbols): d=v{i}t + 2a(t)^2. Rewriting the equation, I had
t= sqrt(d/2a) which gave me 3.11 s, half of their answer. Where did I go wrong? (Or do they have it wrong?)
3 Answers
- remowlmsLv 71 decade agoFavourite answer
t^2 = (2d)/(gf)
t^2 = (2x190)/(9.8*1)
t^2 = 38.7755
t = sqrt(38.7755)
t = 6.2270 sec
Source(s): The original formula you have wrong. It should be d = v(i)t +(a(t^2))/2. Your formula that you rewrote should have been t = sqrt(2d/a). Hope that helps - Cheers! - billrussell42Lv 71 decade ago
d = d₀ + ½at²
a = 9.81 m/s²
d₀ = 0
190 = ½(9.81)t²
t² = 380/9.81 = 38.74
t = 6.22
- Anonymous1 decade ago
ask your teacher. man i hate physics im quiting it soon..
