Related Rate...

[S_A_B]

I'm not quite sure if I'm heading in the right direction or not...so any help would be appreciated...

A ball is dropped from a height of 100ft, at which time it's shadow is 500ft from the ball. How fast is the shadow moving when the ball hits the ground? The ball falls with velocity 32ft/sec, and the shadow is cast by the sun.

Now here is what I've worked out...(and forgive me as I'm not quite fully sure how to post LaTeX code so this may not work... )

\(\displaystyle \frac{{100}}{{500}} = \frac{{\dot y}}{{\dot x}}\)

where:
\(\displaystyle \dot y = 32ft \cdot s^{ - 1}\)

now solving for \(\displaystyle {\dot x}\):

\(\displaystyle \dot x = 5 \cdot \dot y = 5 \cdot 32t = 160t\)

assuming:
\(\displaystyle y = 100 - 16t^2\)

then:
\(\displaystyle {\rm{ }}t = \sqrt {\frac{{100 - y}}{{16}}}\)

Now according to the book the answer is:

\(\displaystyle 40\sqrt {96} ft \cdot s^{ - 1} = 160\sqrt 6 ft \cdot s^{ - 1}\)

Now assuming I didn't do anything wrong...my answer seems correct...if \(\displaystyle y = 16\). But without knowing the answer I wouldn't have been able to get \(\displaystyle y = 16\). In my mind when the ball is at ground level it should be \(\displaystyle y = 0\). So I'm guess I'm asking where I went wrong.

Thank you for any help you can give,
Steven

 

[galactus]

I get 400 ft/sec. Your book says about 392 ft/sec.

Are you sure there is not more to the problem?. Mostly, these problems have the ball being dropped next to a pole or something like that.

Just checking.

 

[S_A_B]

I just double-checked and that is all there is to the problem. Do you think I should just assume the question is in error? Also just to make sure (as I'm not all that confident at these related rate problems) was I going about it the proper way to solve the problem? Thank you very much for your time and help!

 

[soroban]

Hello, S_A_B!

I too got 400 ft/sec.
Then I discovered my mistake . . . *blush*

A ball is dropped from a height of 100 ft, at which time its shadow is 500 ft from the ball.
How fast is the shadow moving when the ball hits the ground?
The ball falls with velocity 32ft/sec, and the shadow is cast by the sun.

      P
   -  *
   :  |   *
   :  |       *
   :  oB          *    500
  100 |   *           *
   :  |       *           *
   :  |h          *           *
   :  |       x       *   S       *
   -  * - - - - - - - -_- o - - - - - *
      R - - - - -  200/6  - - - - - - Q

The ball start at \(\displaystyle P\!:\;PR = 100\)
Its shadow is at \(\displaystyle Q\!:\;PQ = 500\) . (Get it? .The hypotenuse is 500.)
. . Hence: .\(\displaystyle RQ = 200\sqrt{6}\)

\(\displaystyle t\) seconds later, the ball is at \(\displaystyle B:\;h = BR.\)
Its shadow is at \(\displaystyle S\!:\;x = RS\)
\(\displaystyle \text{Since the right triangles remain similar: }\;\frac{x}{h} \:=\:\frac{200\sqrt{6}}{100} \quad\Rightarrow\quad x \:=\:2\sqrt{6}\,h\)

The height of the ball is given by: .\(\displaystyle h \:=\:100-16t^2\)

. . Hence: .\(\displaystyle x \:=\:2\sqrt{6}(100-16t^2)\)

. . \(\displaystyle \text{And: }\:\frac{dx}{dt} \:=\:-64\sqrt{6}\,t\;\;{\bf{[1]}\)


\(\displaystyle \text{When the ball hits the ground, }h = 0\!:\;\;100-16t^2 \:=\:0 \quad\Rightarrow\quad t \,=\,\tfrac{5}{2}\)

\(\displaystyle \text{Substitute into [1]: }\;\frac{dx}{dt} \:=\:-64\sqrt{6}\left(\tfrac{5}{2}\right) \;=\;-160\sqrt{6}\)


Therefore, the shadow is moving at \(\displaystyle 160\sqrt{6}\) ft/sec.

 

[galactus]

Oh, Soroban, I see now. Tricky. The hypotneuse is 500, not x.