Respuesta :
Answer:
The answers to the questions are
Part 1
The terminal speed of the sphere (in m/s) is 44.8973 m/s.
Part 2
The sphere has to be dropped from a height of 102.74 m to reach the speed of 44.8973 m/s if it fell without air resistance.
(102.74 m).
Explanation:
Part 1
Terminal velocity is given by
[tex]v_T = \sqrt{\frac{mg}{\rho C_DA} }[/tex]
Where
[tex]v_T[/tex] = Terminal velocity
m = Mass of the sphere = 3.498 kg
g = Acceleration due to gravity = 9.81 m/s²
ρ = Density of the air = 1.20 kg/m³
[tex]C_D[/tex] = Drag coefficient = 0.500
A = Object cross sectional area = 0.022698 m²
The mass of the sphere is given by
Density = [tex]\frac{Mass}{Volume}[/tex] where volume of the sphere is given by
Volume = [tex]\frac{4}{3} \pi r^3[/tex]
Where
r = radius of the sphere = 8.50 cm = 0.085 m
Therefore the volume V = 2572.44 cm³ = 2.57×10⁻³ m³
The mass then becomes, Mass = Density × Volume = 1.36×2572.44 = 3498.5184 g = 3.498 kg
The mass, m of the sphere = 3.498 kg
The cross sectional area of the sphere is given by A = π×r² = 0.022698 m²
Therefore the terminal velocity becomes
[tex]v_T[/tex] = [tex]\sqrt{\frac{3.498*9.81}{1.20*0.500*0.022698} }[/tex]
= 44.8973 m/s.
Part 2
To reach the speed of 44.8973 m/s in free fall, we have
v² = u² + 2·g·h where
u = initial velocity = 0 m/s
h = Height from which the sphere needs to be dropped
v = final sphere velocity without air resistance
Therefore v² = 2·g·h ⇒ 44.8973² = 2×9.81×h
h = 44.8973²÷(2×9.81)
= 102.74 m.
