Posted by Zac on Thursday, July 26, 2012 at 11:59am.
(a)
In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving. Therefore,
m•g=r•V, where r is drag coefficient,and V is the terminal force.
V= m•g/r.
Now for accelerated motion,
m•a=m•g-r•v,
dv/dt =g –r•v/m,
Δv/Δt= g –r•v/m,
Since Δv =v(fin)-v(initial), and
v(initial) = 0,
v(fin)=V/2 =>
Δv = V/2.
Then
V/2•Δt= g – r•V/2•m,
m•g/2•r• Δt = g - r•m•g/m•r•2
r=m/ Δt =6.7/4.5 = 1.49 kg/s.
V=mg/r= 67•9.8•4.5/6.7=44.1 m/s.
(b) v=3•V/4.
Δv/Δt= g –r•v/m,
3V/4•Δt =g - r•3•V/4•m,
Solve for Δt,
Δt = 3m/r=3•6.7/1.49 =13.49 s.
(c)
m•a = m•g - m•v,
a = g-r•v/m,
Multiply this equation by Δt.
a• Δt =g• Δt - r•v• Δt /m,
Since a• Δt =Δv=V/2 =22.05 m/s, and
v• Δt = Δs, we obtain
Δs =(m/r)(g• Δt – Δv)=
=(6.7/1.49)(9.8•4.5-22.05)=99 m.
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