Answer:
3.5316 m/s²
Explanation:
At terminal velocity v acceleration
a = g
where,
g = Acceleration due to gravity = 9.81 m/s²
[tex]ma=mg-m\left(\frac{4}{5}\right)^2g\\\Rightarrow a=g(1-\frac{16}{25})\\\Rightarrow a=\frac{9}{25}g\\\Rightarrow a=\frac{9}{25}\times 9.81\\\Rightarrow a=3.5316\ m/s^2[/tex]
The acceleration of the skydiver who is currently falling at four-fifths his eventual terminal speed is 3.5316 m/s²
