Respuesta :
Answer: The terminal velocity of the skydiver is 281.30 m/s.
Explanation:
Mass of the sky diver = 75 kg
Density of the air at height of 39,000 m =4.3% of the surface density of the air
= [tex]1.2 kg/m^3\times \frac{4.3}{100}=0.0516 kg/m^3[/tex]
Cross sectional area of the sky diver , A = [tex]0.72 m^2[/tex]
Coefficient of drag ,C = 0.5
The terminal speed is given by the formula:
[tex]v_T=\sqrt{\frac{2mg}{c\rho A}}[/tex]
[tex]v_T=\sqrt{\frac{2\times 75 kg\times 9.8 m/s^2}{0.5\times 0.0516 km/m^3\times 0.72 m^2}}=281.30 m/s[/tex]
The terminal velocity of the skydiver is 281.30 m/s.
The terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
Further Explanation:
When a body falls from the sky it will accelerate continuously in the downward direction by the force of gravity results in increment in the acceleration. Another force is acting on the body is friction force due to air or the particles present in the atmosphere in the in the upward direction or opposite to the direction of motion also called drag force. The drag force is directly proportional to the square of the velocity of falling body. Therefore, with increase in the velocity the drag force also increasing which is proportional to equal to the square of the velocity of falling body. And time will come where the drag force is equal to the weight or the force of gravity results in zero resultant force means zero acceleration. At this point the speed become constant and reached its maximum velocity. This speed is called the terminal velocity.
Given:
The height of the jump is [tex]39000\text{ m}[/tex].
The density air is [tex]4.3\%[/tex] of the surface density.
The density of air at sea level is [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex].
The mass of the skydiver is [tex]75\text{ kg}[/tex].
The cross sectional area of the skydiver is [tex]0.72{\text{ }}{{\text{m}}^2}[/tex].
Concept:
The expression for the terminal speed is:
[tex]\fbox{\begin\\{v_t} =\sqrt {\dfrac{{2mg}}{{\rho A{C_d}}}}\end{minispace}}[/tex] …… (1)
Here, [tex]{v_t}[/tex] is the terminal speed, [tex]m[/tex] is the mass of the skydiver, [tex]g[/tex] is the acceleration due to gravity, [tex]\rho[/tex] is the density, [tex]A[/tex] is the cross sectional area and [tex]{C_d}[/tex] is the drag coefficient.
The term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] and the drag coefficient of a human in free fall is [tex]0.5[/tex].
The density of the air at the altitude of [tex]39000\text{ m}[/tex] is:
[tex]\rho=4.3\%\times{\rho _{air}}[/tex]
Here,[tex]\rho[/tex] is the density of the air at the altitude of [tex]39000\text{ m}[/tex] and [tex]{\rho _{air}}[/tex] is the surface density of the air.
Substitute [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]{\rho _{air}}[/tex] in the above equation.
[tex]\begin{aligned}\rho&=4.3\%\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=\frac{{4.3}}{{100}}\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=0.0516{\text{ kg/}}{{\text{m}}^3}\\\end{aligned}[/tex]
Substitute [tex]0.0516{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]\rho[/tex], [tex]75\text{ kg}[/tex] for [tex]m[/tex], [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex], [tex]0.72{\text{ }}{{\text{m}}^2}[/tex] for [tex]A[/tex] and [tex]0.5[/tex] for [tex]{C_d}[/tex] in equation (1).
[tex]\begin{aligned}{v_t}&=\sqrt{\frac{{2\left({75{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right)}}{{\left({0.0516{\text{ kg/}}{{\text{m}}^3}}\right)\left({0.72{\text{}}{{\text{m}}^2}} \right)\left( {0.5} \right)}}}\\ &=\sqrt{\frac{{1471.5}}{{0.018576}}}{\text{ m/s}}\\&=\sqrt{79215.116}{\text{ m/s}}\\&=281.452{\text{ m/s}}\\ \end{aligned}[/tex]
Therefore, the terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
Learn more:
1. Terminal velocity https://brainly.in/question/6499953
2. Terminal velocity https://brainly.com/question/8789089
3. Net force https://brainly.com/question/11799308
Answer Details:
Grade: High school
Subject: Physics
Chapter: Kinematics
Keywords:
Air, less, dense, higher, elevation, skydivers, reach high, terminal speed, recorded, achieved, jump, height, 39,000 m, density, air, 4.3%, surface density, estimate, sea, level, , mass, 75 kg, cross section, area, , 281.452 m/s.
