A 22 cm diameter bowling ball has a terminal speed of 85 m/s. What is the ball's mass?

Answer 1

Vterminal=square root of (4mg/pa)
p=1.22(density of air)
a=22cm=.22m/2=.11m(radius)... (pi)r^2=.038=a
85^2 = 4mg/pa
7225 = 4(m)(9.8)/(1.22)(.038)
7225(.038)(1.220)/(4)(9.8) = m
8.54kg=m

Answer 2

Vterminal=square root of (4mg/pa)
p=1.22(density of air)
a=22cm=.22m/2=.11m(radius)... (pi)r^2=.038=a
85^2 = 4mg/pa
7225 = 4(m)(9.8)/(1.22)(.038)
7225(.038)(1.220)/(4)(9.8) = m
8.54kg=m

Answer 3

Vterminal=square root of (4mg/pa)
p=1.22(density of air)
a=22cm=.22m/2=.11m(radius)... (pi)r^2=.038=a
85^2 = 4mg/pa
7225 = 4(m)(9.8)/(1.22)(.038)
7225(.038)(1.220)/(4)(9.8) = m
8.54kg=m