Answer:
115 m/s, 414 km/hr
Explanation:
There are two forces acting on a skydiver: gravity and air resistance (drag). At terminal velocity, the two forces are equal and opposite.
∑F = ma
D − mg = 0
D = mg
Drag force is defined as:
D = ½ ρ v² C A
where ρ is the fluid density,
v is the velocity,
C is the drag coefficient,
and A is the cross sectional surface area.
Substituting and solving for v:
½ ρ v² C A = mg
v² = 2mg / (ρCA)
v = √(2mg / (ρCA))
We're given values for m and A, and we know the value of g. We need to look up ρ and C.
Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.
For a skydiver falling headfirst, C ≈ 0.7.
Substituting all values:
v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))
v = 115 m/s
v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)
v = 414 km/hr
The terminal velocity of the skydiver in m/s and km/h is; 115m/s and 414 km/h
Using Given data :
mass of skydiver ( M ) = 80 kg
Cross sectional surface area ( A ) = 0.14 m^2
p ( fluid density ) ≈ 1.21 kg/m³.
C ( drag coefficient ) = 0.7
Determine the terminal velocity of the skydiver
At terminal velocity drag force and gravity is equal and opposite therefore canceling out each other
∑ F = ma
Drag force - Mg = 0
therefore; D = Mg ----- ( 1 )
where D ( drag force ) = 1/2 pv² C A ---- ( 2 )
p = fluid density , C = drag coefficient , A = cross sectional area
v = velocity
Back to equations 1 and 2 ( equating them )
1/2 pv² CA = Mg ---- ( 3 )
v² = 2mg / ( p C A )
∴ V = √ ( 2mg / (p C A ))
V = √ ( 2 * 80 * 9.8 ) / ( 1.21 * 0.7 * 0.140 ))
= 115 m/s
also V = 414 km/h
Hence we can conclude that the terminal velocity of the skydiver is in m/s and km/h are 115m/s and 414 km/h
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