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Calculate the velocity a spherical raindrop would achieve falling (take downward as positive) from 4.2 km in the following situations.

(a) Calculate the velocity in the absence of air drag.

(b) Calculate the velocity with air drag. Take the size across of the drop to be 4.8 mm, the density of air to be 1.17 kg/m3, the density of water to be 1000 kg/m3, the surface area to be πr2, and the drag coefficient to be 1.0.

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Answer:

287.06 m/s

7.2 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s

r = Radius of drop

[tex]\rho[/tex] = Density

[tex]v^2-u^2=2gs\\\Rightarrow v=\sqrt{2gs+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 4200+0^2}\\\Rightarrow v=287.06\ m/s[/tex]

Velocity without air drag = 287.06 m/s

[tex]v=\sqrt{\frac{2mg}{\rho CA}}[/tex]

[tex]m=\rho_w V=\rho_w\frac{4}{3}\pi r^2\\\Rightarrow m=1000\times \frac{4}{3}\times \pi \times 0.0024^3[/tex]

[tex]v=\sqrt{\frac{2\times 1000\times \frac{4}{3}\times \pi \times 0.0024^3\times 9.81}{1.21\times 1\times \pi \times 0.0024^2}}\\\Rightarrow v=7.20\ m/s[/tex]

Velocity with air drag = 7.20 m/s

(a) The change of displacement with respect to time is defined as the velocity. Velocity in the absence of air drag will be 287.06 m/s.

(b) The velocity with air drag will be 7.2 m/s.

What is velocity?

The change of displacement with respect to time is defined as the velocity.  

velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

The given data in the problem is given as;

t is the time taken

u is the  initial velocity=0

v is the final velocity

s is the displacement

g is the acceleration due to gravity = 9.81 m/sÂČ

r is the radius of the drop =

ρₐ is the density of water =1000 kg/mÂł

ρₓ is the density of air=1.17 kg/mÂł

c is the coefficient of drag= 1

(a) The velocity in the absence of air drag will be 287.06 m/s.

According to newton's third equation of motion

[tex]\rm v^2=u^2+2as\\\\\rm v^2=2gs\\\\\rm v=\sqrt{2gs} \\\\\rm v=\sqrt{2\times9.81\times4200}\\\\\rm v=287.06 \;m/sec[/tex]

Hence the velocity in the absence of air drag will be 287.06 m/s.

(b) The velocity with air drag will be 7.2 m/s.

The formula for velocity with the drag is given by

[tex]\rm m=\rho V\\\\\rm m=\rho \frac{4}{3}\pi r^2 \\\\\rm m=\rho V\\\\\rm m=1000 \times \frac{4}{3}\3.14 (0.0024)^2 \\\\\rm m=5.790\times10^{-5}\\\\\\\\\rm A=\pi r^2\\\\ \rm A=3.14\times (0.0024)^2\\\\ \rm A=1.80\times 10^{-5}\\\\\\\\ \rm v_d=\sqrt{\frac{2mg}{\rho CA} } \\\\\ \rm v_d=\sqrt{\frac{2\times5.70\times10^{-5}\times9.81}{1.21\times1\times1.80\times10^{-5}}[/tex]

[tex]\rm v_d=7.5 \;m/sec.[/tex]

Hence the velocity with air drag will be 7.2 m/s.

To learn more about the velocity refer to the link;

https://brainly.com/question/862972

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