Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be 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.
Trending News
This should be simple...?
I'm taking a basic physics class after several years of not taking any kind of math or physics classes. This question should be incredibly simple, but I'm absolutely stuck for reasons I can't understand. I never had any problems with advanced math, but I'm stumped now.
how long would it take an object to fall 2 m? We can use 10m/s2.
I see time, acceleration, and distance being the variables, but for the life of me I can't put them together, and my book doesn't give me any help.
How do you work this out?
2 Answers
- AnthonyLv 79 years agoFavorite Answer
Use s = ut + 1/2 at^2
where u = initial velocity = 0, a = acceleration = 10m/s^2, t = time in seconds s = distance in metres = 2m.
2 = 0 + 5t^2
2/5 = t^2
0.4 = t^2
0.632s = t
- dawgdaysLv 79 years ago
It's algebra. You can start with the equation
D = D(initial) + V(initial)*t + 0.5*a*t^2
Since V(initial) is zero and D(initial) is zero at the drop point, and the a is -10 m/s^2 (down is negative by convention), you end up with
-2 = -10*t^2
2/10 = 1/5 = t^2
t = sqrt(1/5)
Make sense?