A steel ball of mass 4-kg is dropped from rest

Donald Johnson

Donald Johnson

Answered question

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.

Answer & Explanation

Piosellisf

Piosellisf

Beginner2021-12-30Added 40 answers

Mass of the ball m=4kg
Dropped from nest do initial velocity is zero is v=0 when t=0
Given that air resistence =0.012v
The equation of motion is
mdvdt=g0.012v
or dvdt=g0.012vm
=g0.012v4
dvg0.012v=dt4
or d(g0.012v)0.012(g0.012v)=dt4
or d(g0.012v)g0.012v=dt4
=0.003dt
Integrating d(g0.012v)g0.012v=0.003dt+c
or ln(g0.012v)=0.003t+C
Initially v=0,when t=0
PSKln(g)=c
ln(g0.012v)=0.003t+lng
ln(g0.012vg)=0.003t
g0.012vg=e0.003t
10.012vg=e0.003t
1e0.003t=0.012gv
v=g0.012(1e0.003t)
dxdt=g0.012(1e0.003t)
dx=g0.012(1e0.003t)dt
Integrating,
raefx88y

raefx88y

Beginner2021-12-31Added 26 answers

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
=33.32ms
Resistance =0.015v
=0.015×(33.32)
=0.5ms2
h=u+y2g+2 NSk h=0+12(9.80.5)×3.42
h=53.754m
Height of the building is 53.754m
karton

karton

Expert2022-01-09Added 613 answers

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
a=gf/m=9.80.031v/4=9.83.25103vAlso we know that a=dv/dtdv/dt=9.83.25103,Integrating both sides0vdvgbv=0tdt(b=3.25103)1bln(gbv)=tgbv=ebtv=gebtb=9.80.9883.25103=2.71103m/sAlso v=dxdt=gebtbx=g+ebt/bb=9.8+304.013.25103=95.67km

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