Answer:
The terminal speed is 74.833 m/s
Explanation:
The drag force is equal to square of speed:
Fdrag = k*v²
According Newton`s law:
Fnet = m*a
m*g - k*v² = m*a
[tex]k=\frac{m(g-a)}{v^{2} }[/tex]
[tex]k=\frac{16*(9.8-4.6)}{54.2^{2} } =0.028[/tex]
If terminal speed, the net force is zero.
[tex]kv_{t} ^{2} =mg\\v_{t} =\sqrt{\frac{mg}{k} } =\sqrt{\frac{16*9.8}{0.028} } =74.833m/s[/tex]