Question

A rock weighing 4lbs is dropped from rest. Air resistance is -1/2v. Find v(t) and s(t) at time, t=5 seconds.

Answer 1

For body of mass m, if air resistance is -kv

Here m=4lbs, k=1/2, g=32.2 feet/s²(in fps system)

v(t) = (mg/k)(1 - e^(-k/m)t)

=4*32.2*2* [1 - e^(-1/8)t]

=258*[1 - e^-5/8]

=258*[1-0.54] = 120 feet/s

By integrating the velocity by time, we can get below distance equation

s(t)= (mg/k)(t + m/k(e^(-k/m)t - 1) )

=258(5+ 8(e^-5/8-1) ) =258 (5 - 3.68)= 258*1.32= 340 feet