Answer:
The terminal velocity of the diver is 115 m/s = 414 km/hr
Explanation:
At terminal velocity,
Fnet = mg - Fd = 0
Drag force, Fd = cρAv²/2
mg = cρAv²/2
Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by
v = √(2mg/ρcA)
where m = mass of body falling through fluid = 80 kg
g = acceleration due to gravity = 9.8 m/s²
ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³
c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7
A = the area of the diver facing the fluid = 0.14 m²
v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr
Speed of any freely falling object is always same. Provided, both are left to fall from the same height. If you perform this experiment in a perfect vacuum or near vacuum laboratory, both of them will reach ground with same velocity this is because there is no resistance to their motion. This is always true no matter where you go and perform this experiment.
It can be easily proved from conservation of mechanical energy. Why conserving energy? because there are no forces acting on the freely falling objects other than conservative force(mg).