How about this one?
Q: A girl drops a coin from the roof top to the ground 6 meters below. If the coin is released from rest, how long is it in the air?
t = ?
h = 6 meters
2007-07-15 07:04:03 · 4 answers · asked by mikko 1 in Science & Mathematics ➔ Physics
h(distance from ground) = 6m
a(acceleration) = 9.81m/s²
u(inital speed) = 0m/s
Use s = ut + ½at² where s = h
Hence h = ut + ½at²
We know u = 0 hence:
h = ½at²
Rearrange to solve for t:
h = ½at²
2h = at²
2h/a = t²
√(2h/a) = t
Hence putting the numbers in:
√(2 x 6/9.81)= t
t = 1.1 secs.
2007-07-15 08:00:34 · answer #1 · answered by Tsumego 5 · 0⤊ 0⤋
Use h = (1/2)*g*t^2
2007-07-15 10:56:25 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
If the s(h)=g/2 x t*2 means that "t" is square root from 6/5 or square root from 1.2. Result is 1.095 seconds. Important is that girl is just let go coin, not THROW him (BECAUSE IN THIS CASE YOU HAVE TO COUNT AND FORCE (F) OF THROW. Veradisca & Best Regards, Neven.
2007-07-15 07:13:36 · answer #3 · answered by NEVEN , 4 · 0⤊ 0⤋
h = 1/2 * g * t^2
h = 6 m
g = 9.8 m/s^2
6 = .5 * 9.8 * t^2
t = sqrt(1.22)
t = 1.1 sec
2007-07-15 07:08:04 · answer #4 · answered by Anonymous · 0⤊ 0⤋