Calculus and Baseball

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November 16th 2009, 08:17 AM
ReneePatt

Calculus and Baseball

An outfielder fields a baseball 280 ft away from home plate and throws it directly to the cather with an initial velocity of 100 ft/s. Assume that the velocity v(t) of the ball after t seconds satisfies the differential equation dv/dt = -1/10v because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball).

I apologize if this question has already been posted. But I've spent some time searching and haven't found anything yet.
November 16th 2009, 10:07 AM
TKHunny

The simple models are all the same. Somehow, cram it in your head.

$s(t) = \frac{1}{2}at^{2} + v_{0}t + h_{0}$

As long as we have only one component of acceleration, this is all we need to know. I'll ignore the units. You make sure everything works.

Quote:

Ignore any vertical motion of the ball

$h_{0} = 0$

$s(t) = \frac{1}{2}at^{2} + v_{0}t$

Quote:

an initial velocity of 100 ft/s

$v_{0} = 100$

$s(t) = \frac{1}{2}at^{2} + 100t$

Quote:

dv/dt = -1/10v

$a = -\frac{1}{10}$

$s(t) = \frac{1}{2}\left(\frac{-1}{10}\right)t^{2} + 100t$

Now what?

Hint: The solution is NOT 2000 seconds. You tell me why.
November 16th 2009, 12:42 PM
topsquark

Quote:

Originally Posted by TKHunny

$a = -\frac{1}{10}$

Careful!

I have to ask. There are two possible interpretations of what the OP wrote. To the OP: Is the acceleration
$a = -\frac{v}{10}$ (as it is written)
or is it
$a = -\frac{1}{10v}$ (as many students might write it to be)

If it is the first case then
$a = -\frac{v}{10}$

$\frac{dv}{dt} = -\frac{v}{10}$

$\frac{dv}{dt} + \frac{1}{10}v = 0$

Then v(t) must be of the form
$v = Ae^{-t/10}$

Now apply your initial conditions and you are done.

If the equation is the second of the two, that's a much more involved solution. If you need that solved, just say so and someone will help you.

-Dan
December 3rd 2009, 09:29 AM
ReneePatt

Calculus & Baseball - Part 2

Thanks
December 9th 2009, 06:21 PM
ReneePatt

Does this look correct?

This is how I worked the problem out and what I got - is it correct?

h'(t) = -32t + c

$v_0$ =h'(0)=-32(0)+c=c

h'(t)= $v_0$ -32t

h(0)=0

h(t)= $v_0{t}-16t^2$

h'(t)=0

0=h'(t)= $v_0$ -32t

$t=\frac{v_0}{32}$

$h(\frac{v_0}{32})=v_0(\frac{v_0}{32})-16(\frac{v_0}{32})^2=\frac{v_0^2}{32}-\frac{v_0^2}{64}=\frac{v_0^2}{64}$

$280=\frac{v_0^2}{64}$

$17920=v_0^2$

$\sqrt{17920}=v_0=133.87\ sec$
December 12th 2009, 09:25 AM
ReneePatt

Quote:

Originally Posted by TKHunny

The simple models are all the same. Somehow, cram it in your head.

$s(t) = \frac{1}{2}at^{2} + v_{0}t + h_{0}$

Where did you get this equation?

Quote:

Originally Posted by TKHunny

$s(t) = \frac{1}{2}\left(\frac{-1}{10}\right)t^{2} + 100t$

Now what?

Hint: The solution is NOT 2000 seconds. You tell me why.

I'm not sure what to do next. How would you solve for this and NOT get 2000 seconds because that's exactly what I got?
December 13th 2009, 07:06 AM
HallsofIvy

I'm completely confused! There seem to be two different problems being worked on here.

For the problem I saw posted, horizontal motion with acceleration (-1/10)v, topsquarks exponential function is correct. (I did it thinking the acceleration was -1/(10v) and got a square root function.) But your last post and TKHunny's seem to be about an object dropping under acceleration due to gravity.
December 13th 2009, 09:27 AM
ReneePatt

Quote:

Originally Posted by HallsofIvy

I'm completely confused!

You and me both. I have no clue how to work this problem out and am not understanding any of the suggestions. This is due tomorrow and it looks like it'll go unanswered. I'll just have to show the different ways I tried to solve the problem and see if that gets me any where.