Donald Johnson

2021-12-29

A steel ball of mass 4-kg is dropped from rest from the top of a building. If the air resistance is 0.012v and the ball hits the ground after 2.1 seconds, how tall is the building? Answer in four decimal places.

Piosellisf

Mass of the ball $m=4kg$
Dropped from nest do initial velocity is zero is
Given that air resistence $=0.012v$
The equation of motion is
$m\frac{dv}{dt}=g-0.012v$
or $\frac{dv}{dt}=\frac{g-0.012v}{m}$
$=\frac{g-0.012v}{4}$
$\frac{dv}{g-0.012v}=\frac{dt}{4}$
or $\frac{d\left(g-0.012v\right)}{-0.012\left(g-0.012v\right)}=\frac{dt}{4}$
or $\frac{d\left(g-0.012v\right)}{g-0.012v}=\frac{dt}{4}$
$=-0.003dt$
Integrating $\int \frac{d\left(g-0.012v\right)}{g-0.012v}=-0.003\int dt+c$
or $\mathrm{ln}\left(g-0.012v\right)=-0.003t+C$
Initially
$\therefore PSK\mathrm{ln}\left(g\right)=c$
$\therefore \mathrm{ln}\left(g-0.012v\right)=-0.003t+\mathrm{ln}g$
$\mathrm{ln}\left(\frac{g-0.012v}{g}\right)=-0.003t$
$\frac{g-0.012v}{g}={e}^{-0.003t}$
$1-\frac{0.012v}{g}={e}^{-0.003t}$
$1-{e}^{-0.003t}=\frac{0.012}{g}v$
$v=\frac{g}{0.012}\left(1-{e}^{-0.003t}\right)$
$\frac{dx}{dt}=\frac{g}{0.012}\left(1-{e}^{-0.003t}\right)$
$dx=\frac{g}{0.012}\left(1-{e}^{-0.003t}\right)dt$
Integrating,

raefx88y

Solution:
Given, mass of the ball, $m=4kg$
air resistance, $=0.015v$
$t=3.4$ seconds
Now, speed at the ball at the bottom
$v=u+>$
$=0+9.8×3.4$
$=33.32\frac{m}{s}$
Resistance $=0.015v$
$=0.015×\left(33.32\right)$
$=0.5\frac{m}{{s}^{2}}$
$h=u+{y}_{2}g+2$ NSk $h=0+\frac{1}{2}\left(9.8-0.5\right)×{3.4}^{2}$
$h=53.754m$
Height of the building is 53.754m

karton

Mass of the body m=4 kg, time taken to hit the ground t=3.7s and air resistance f=0.013v. Since the ball was dropped, the initial velocity u=0.
Force equation for the ball, F=ma=mg-f

Do you have a similar question?