For the
presented problem, the solution would be v(0)=0v(0)=0 isv(t)−mgb=e−b/m⋅t(v0−mgb)⟺v(t)=mgb(1−e−b/m⋅t)≈g⋅t−gb2m⋅t2,
with the following given,
m=180[lb]=81.6[kg]
g=9.81[m/s2]
b=0.75[kg⋅m/s2⋅s/ft]=0.2286[kg/s]
The solution that the
friction provides is v(t)=3501.7[m/s]⋅(1−e−0.00280[1/s]⋅t), where I get 96.69[m/s]=317.2[ft/s]. I
am hoping that this answer has satisfied your query about this specific
question.
