Shadow Problem

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jschwa1

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Aug 2, 2010
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A light is on top of a pole 25 ft above ground. A ball is raised to the same height and is 35 ft away from the light. If the ball is dropped, at what elevation is its shadow moving at 500 ft/sec.

There is no diagram with this problem and I'm having a tought time figuring out an equation to relate the variables. Thanks.
 
There are various ways to tackle this problem.

I hope I got it correct :wink:

The ball has position \(\displaystyle y=25-16t^{2}\).................[1]

Thus, \(\displaystyle \frac{dy}{dt}=-32t\)

By similar triangles:

\(\displaystyle \frac{25}{y}=\frac{35+x}{x}\)

\(\displaystyle x=\frac{35y}{25-y}\)

Sub in [1]: \(\displaystyle x=\frac{35(25-16t^{2})}{25-(25-16t^{2})}\)

\(\displaystyle \frac{dx}{dt}=\frac{875}{8t^{3}}\)

We know that \(\displaystyle \frac{dx}{dt}=500\)

\(\displaystyle 500=\frac{875}{8t^{3}}\)

\(\displaystyle t=\frac{\sqrt[3]{14}}{4}\approx .6025 \;\ sec\)

At this time, the ball is at height:

\(\displaystyle y=25-16\left(\frac{\sqrt[3]{14}}{4}\right)^{2}=25-\sqrt[3]{196}\approx 19.2 \;\ ft\)

Note, it only takes 5/4 seconds for the ball to hit the ground.
 
galactus, thanks so much for your help. I'm still a bit confused because I'm getting dx/dt = -875/t^3. Am I doing something wrong?
 
Perhaps I should have used -500 to explain. That is because as the ball drops, the shadow moves inward toward the pole.

I just shed the negative signs.

Maybe Soroban (who likes these kinds of problems), or someone else, will be around after while to either confirm or deny my solution.
 
Hello, jschwa1!

My interpretation differs from Galactus' . . . but we arrive at the same answer!

A light is on top of a pole 25 ft above ground.
A ball is raised to the same height and is 35 ft away from the light.
If the ball is dropped, at what elevation is its shadow moving at 500 ft/sec.
Click to expand...
Code: 
      L
      o
      |     *     B
      |           o
   25 |           |     *
      |           |y          *
      |           |               *
      *-----------*---------------------o
      : -  35 - - : - - - x-35  - - - - S
      : - - - - - - - x - - - - - - - - :

\(\displaystyle \text{The light is at }L.\)
. . \(\displaystyle \text{Its height is 25 feet.}\)

\(\displaystyle \text{The ball is at }B.\)
. . \(\displaystyle \text{Its height is: }\,y \:=\:25-16t^2\,\text{ feet.}\)

\(\displaystyle \text{The shadow is at }S.\)
. . \(\displaystyle \text{Let }x\text{ = distance of }S\:f\!rom\;the\;lightpole.\)


\(\displaystyle \text{From similar triangles, we have: }\:\frac{y}{x-35} \:=\:\frac{25}{x} \quad\Rightarrow\quad \frac{25-16t^2}{x-35} \:=\:\frac{25}{x}\)

\(\displaystyle \text{This simplifies to: }\:x \:=\:\frac{875}{16t^2} \quad\Rightarrow\quad x \:=\:\frac{875}{16}t^{-2}\)

\(\displaystyle \text{Differentiate with respect to time: }\:\frac{dx}{dt} \:=\:-\frac{875}{8}t^{-3} \quad\Rightarrow\quad \frac{dx}{dt}\:=\:-\frac{875}{8t^3}\)


\(\displaystyle \text{The shadow is moving at 500 ft/sec: }\:\frac{dx}{dt} \,=\,-500\)

. . \(\displaystyle \text{So we have: }\;-500 \:=\:-\frac{875}{8t^3} \quad\Rightarrow\quad t^3 \:=\:\frac{875}{4000} \:=\:\frac{14}{64}\)

\(\displaystyle \text{Hence: }\:t \;=\;\sqrt[3]{\frac{14}{64}} \:=\:\frac{\sqrt[3]{14}}{4} \;\approx\;0.6025\text{ sec. }\;\hdots\;\text{ etc.}\)

 
Cool, Soroban. Our interpretations were close. At least we arrived at the same answer.

Maybe this will help jschwa1 with other related rates problems and how to set them up.

They do not give the 'physics' part of the problem with regards to a falling object.

Which is \(\displaystyle y=-\frac{1}{2}gt^{2}+v_{0}t+y_{0}\).
 
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