Respuesta :
Answer:
The displacement of the air drop after 3 second is 18.27 m.
Explanation:
Mass of the rain drop = m = [tex]0.5\times 10^{-4} kg[/tex]
Weight of the rain drop = W
Duration of time = t = 3 seconds
[tex]W=m\times g[/tex]
Drag force on rain drop = [tex]D=0.2\times 10^{-5} v^2[/tex]
[tex]W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N[/tex]
Motion of the rain drop:
[tex]F=m\times a[/tex]
Net force on the rain drop , F= Â W - D
[tex]W-D=m\times a[/tex]
[tex]0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a[/tex]
[tex]0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}[/tex]
[tex]0.006v^2+0.05v-1.5=0[/tex]
v = 12.18 m/s
Initial velocity of the rain drop = u = 0 (since, it is starting from rest)
v=u+at (First equation of motion)
[tex]12.18 m/s=0m/s+a\times 3 s[/tex]
[tex]a=4.06 m/s^2[/tex]
[tex]s=ut+\frac{1}{2}at^2[/tex] (second equation of motion)
[tex]s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2[/tex]
s = 18.27 m
The displacement of the air drop after 3 second is 18.27 m.
