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

Answer 2

If I'm not wrong :)

Answer 3

the solution above is incorrect. It's clear because the acceleration due to gravity for any object is 9.8 m/s^2, so any object in free-fall will have a maximum velocity of 9.8 m/s after 1 second. Taking into account the drag (air resistance), it will be less than 9.8 m/s after that 1 second. So an answer of 24 m/s after 1 sec is impossible.

Using the second law, we have:

\(\sum F_y=F_g+D=ma\)

\(\sum F_y=-mg+(\dfrac{1}{2}C\rho Av^2)=ma\)

\(a=\dfrac{-mg+(\dfrac{1}{2}C\rho Av^2)}{m}\)

Then we'd use the acceleration into the kinematic equation: \(\sf v_f=v_i+a\Delta t\)

\(v_f=v_i+\dfrac{-mg+(\dfrac{1}{2}C\rho Av^2)}{m}\Delta t\)

\(\Delta t=t_f-t_i\), \(t_i=0\), so \(\Delta t=t_f=t\)

\(v_i=0\)

so, \(v_f=-gt+\dfrac{C\rho Av^2 t}{2m}\)

I'm not sure if the drag force would be zero here because the initial velocity is zero, or you would need an integral.