Respuesta :
Answer:
a) [tex]v=313.209\ m.s^{-1}[/tex]
b) [tex]v_t=3751.79\ m.s^{-1}[/tex]
Explanation:
Given:
- height of the raindrop, [tex]h=5000\ m[/tex]
a)
[tex]v=\sqrt{2g.h}[/tex]
[tex]v=\sqrt{2\times 9.81\times 5000}[/tex]
[tex]v=313.209\ m.s^{-1}[/tex]
b)
given that:
diameter of the drop, [tex]d=4\ mm=0.004\ m[/tex]
density of the air, [tex]\rho=1.18\ kg.m^{-3}[/tex]
the terminal velocity is given as:
[tex]v_t=\sqrt{\frac{2m.g}{\rho.A.c_d} }[/tex]
where:
m = mass
g = acceleration due to gravity
[tex]\rho=[/tex] density of the medium through which the drop is falling (here air)
A = area normal to the velocity of fall
[tex]c_d=[/tex] coefficient of drag = 0.47 for spherical body
[tex]v_t=\sqrt{\frac{2\times 5\times 9.81}{1.18\times \pi\times 0.002^2\times 0.47} }[/tex]
[tex]v_t=3751.79\ m.s^{-1}[/tex]
