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maxonik [38]
3 years ago
11

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

s are accurate to three significant digits. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.) If each skydiver has a frontal area of 0.14 m2, calculate their terminal velocities (in m/s). 52 kg skydiver m/s 95 kg skydiver m/s How long will it take (in s) for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)
Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J

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Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm

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