Question

A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k=1/128 lb-s/ft. Find the maximum height attained by the stone.

Answer 1

@SithsAndGiggles

Answer 2

As before, we have the same setup:
\[\begin{align*}
mg+kv&=m\frac{dv}{dt}\\
\left(\frac{1}{2}\right)\left(-32\right)-\left(\frac{1}{128}\right)v&=\frac{1}{2}\frac{dv}{dt}\\
\frac{dv}{dt}+\frac{1}{64}v&=-32
\end{align*}\]
and you're given the initial velocity, so \(v(0)=32\).

Additionally, you're given the initial height, so you can integrate the velocity function and use the initial condition \(h(0)=5\), where \(v(t)=h'(t)\). Alternatively, you can determine when \(v(t)=0\).

Answer 3

So I got\[v=-2048+2080e ^{-\frac{ t }{ 64 }}\]

Answer 4

Since v(t)=h'(t), I integrated v(t) and got
\[h(t)=-2048t-133120e ^{-\frac{ t }{ 64 }}+133125\]

Answer 5

@SithsAndGiggles

Answer 6

mg+kv=m(dv/dt)
since W=mg=-1/2 because it's downwards,
-1/2-(1/128)v=(1/64)(dv/dt)
since you said g=32, I got mass=1/64
dv/dt+(1/2)v=-32
integrating factor
e^(t/2)
e^(t/2)v'+e^(t/2)/2*v=-32e^(t/2)
e^(t/2)v=-64e^(t/2)+C
v=-64+Ce^(-t/2)
v(0)=32
C=98
v=-64+98e^(-t/2)
v(t)=h'(t)
h'(t)=-64+98e^(-t/2)
integrate h'(t)
h(t)=-64t-196e^(-t/2)+C
h(0)=5
C=201
h(t)=-64t-196e^(-t/2)+201
v(t)=0 when t=0.852169
h(0.852169)=18.4612
But the answer in the book is 17.10 ft.
This problem is driving me crazy. What's wrong?

Answer 7

Oh hold up, looks like \(C=96\), not \(98\).

Answer 8

\[32=-64+Ce^{-0/2}~~\implies~~96=C\]

Answer 9

HAHAHAHAHA, I know, right? I'm so sorry about that. Thank you so much for pointing out the error. You're so good.

Answer 10

YES! I finally got the correct answer! Thank you so much for pointing out the error.

Answer 11

You're welcome!