Respuesta :
Answer:
The terminal speed of his is 137.68 m/s.
Explanation:
Given that,
Mass of skydiver = 78 kg
Area of box[tex]A =24\times35=840\ cm[/tex]
Drag coefficient = 0.80
Density of air [tex]\rho= 1.2\times kg/m^3[/tex]
We need to calculate the terminal velocity
Using formula of drag force
[tex]F_{d} = \dfrac{1}{2}\rho v^2Ac[/tex]
Where,
[tex]\rho[/tex] = density of air
A = area
C= coefficient of drag
Put the value into the formula
[tex]78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80[/tex]
[tex]v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}[/tex]
[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}[/tex]
[tex]v=137.68\ m/s[/tex]
Hence, The terminal speed of his is 137.68 m/s.
The terminal speed of the skydiver with dimensions as a rectangular box as he falls feet first is 137.68 m/s.
What is the terminal speed?
Terminal speed of a body is the maximum speed, which is achieved by the object when it fall through a fluid.
In the case of terminal velocity, the force of gravity becomes equal to the sum of the drag force and buoyancy force due to fluid on body.
Terminal velocity can be find out as,
[tex]v=\sqrt{\dfrac{2mg}{\rho AC_d}}[/tex]
Here, (m) is the mass, (g) is gravitational force, ([tex]\rho[/tex]) is the density of fluid, (A) is the project area and ([tex]C_d[/tex]) is the drag coefficients.
It is given that, the mass of the skydiver is 78 kg The dimensions of the skydiver is s 24 cm × 35 cm × 170 cm.
The coefficient of drag is 0.80 and the density of air is 1.2 kg/m³.
Put the values in the above formula as,
[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times0.24\times0.35\times 0.8}}\\v=137.68\rm m/s[/tex]
Thus the terminal speed of the skydiver as he falls feet first is 137.68 m/s.
Learn more about the terminal speed here;
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