Question

If a cotton ball is dropped from 12 meters with air resistance, what will be the velocity and acceleration at t = 1.00 s?

Answer 1

applying F.P.D (Newton's 2nd Law)
(summation)Forces/cotton-ball = m.a (vectors over the Forces and a)
where a is the acceleration

W(vector) + R(vector) = m.a(vector)
R is the air resistance
since a cotton ball is massless then W = m.g = 0 x 10 = 0 N
projection along direction of motion (vertically download)
Rcos(pi) = m.a
-R = m.a
but again the cotton ball is massless then -R = 0 then R = 0 !!

x = (1/2)a.t^2 + V(0)t + x(0)
12 = (1/2)a.(1)^2 + 0(1) + 0
a = 24 m/s^2

V = (integral) a dt = 24t + V(0) = 24t + 0 = 24t

V(1) = 24 m/s^2 :p

Answer 2

Thank you!