This is a hard one!

D

drawrof

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Jul 6, 2005
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I was assigned this problem for homework but the teacher didn't have enough to go over it. Can someone PLEASE help. I just want to know if i got the right answer.

Problem:
A ball is dropped from a height of 20 meters, 12 meters away from the top of a 20-meter lamppost. The ball's shadow. caused by the light at the top of the lamppost, is moving along the level ground. How fast is the shadow moving 1 second after the ball is released?
 
Make the Similar Triangles work for you.

Draw a picture.

1) Start with a rectangle, taller than wide.
2) Label the height 20m
3) Label the width 12m
4) Decide which vertical side is the lamp post - label it.
5) The other side must be the path of the ball - label it.
6) Now, and this it the exciting part, half way down the ball path, draw in the ball - a representation of where it will be at some moment during its descent.
7) Draw a line from the lamp to the ball and keep drawing the line until it reaches ground level.
8) Extend the rectangle base out to meet the line you just created.

OK, now let's label a few points, so we can talk about it.

Label the lamp, A
Label the base of the lamp, B
Label the representative location of the ball, C
Label where the ball will land, D
Label where AC extended and BD extended meet at the ground, E, the shadow of the ball.

Note a couple of things:

m(Angle(ABE)) = m(Angle(CDE)) = 90°
m(Angle(BAE)) = m(Angle(DCE))
And with the reflexive congruence at the other angle, this makes similar Right triangles, Triangle(ABE) and Triangle(CDE)

Once you get a good drawing, it should be more obvious how to proceed. Do you remember your formula for a free-falling object dropped from a given height? You'll need that.
 
Private email motivates additional response:

The basic equation for the location of a free-falling object.

f(t) = ½*g*t<sup>2</sup> + v<sub>0</sub>*t + h<sub>0</sub>

g = Acceleration due to Earth's Assumed Uniform Gravitational Effect = -9.81 m/s<sup>2</sup>

t = Time in seconds

v<sub>0</sub> = Initial Velocity - If you just drop it, v<sub>0</sub> = 0 m/s

h<sub>0</sub> = Initial height - In this case, h<sub>0</sub> = 20 m

So,
f(t) = -4.905 m/s<sup>2</sup>*t<sup>2</sup> + 20 m

f(0) = 20 m -- Exactly as hoped. Where is the shadow at t = 0?
f(2.019275109) = 0 m -- Good. It landed. Where is the shadow now?

OK, let's see where you can get from this...
 
Similar Triangles Suggests:

BE/20 = DE/f(t)
Note: Observe that this does NOT work for f(0) = 20 OR for f(t<sub>splat</sub>) = 0. You tell me why this is so and why it is important.

The nature of the problem indicates BE and DE are also functions of t. this gives:

BE(t)/20 = DE(t)/f(t)

Further, the drawing suggests that BE = 12 + DE, giving:

[12+DE(t)]/20 = DE(t)/f(t)

We know f(t). Solve for DE(t) and give it your best to simplify.

Are we getting closer?
 
It is better for other students if you post your question here, rather than PM.

You're going to make me do the whole thing? Picking up where we left off...

DE(t) = (1/20)*s(t)*[DE(t)+12]
20*DE(t) = s(t)*DE(t) + 12*s(t)
DE(t)*[20-s(t)] = 12*s(t)

Note: 20 - s(t) = 20 - (½g*t<sup>2</sup>+20) = -½g*t<sup>2</sup>

DE(t) = 12*s(t)/[-½g*t<sup>2</sup>] = 12*[½g*t<sup>2</sup>+20]/[-½g*t<sup>2</sup>]
DE(t) = 12 + 480/[g*t<sup>2</sup>]

Now what? We are SO CLOSE!
 
next would i look for the derivative of DE(t)? or would i look for DE(t) where t=1?

I'm obviously having a hard time with this problem. =(

This problem is way overdue but I still want to figure it out. It's bothering me that I can't do it on my own. lol. Thanks so much for helping me.
 
Let's review:

How fast is the shadow moving 1 second after the ball is released?
Click to expand...

That would be DE'(t) evaluated at t = 1.

Well, what's the derivative?
 
the derivative would be -36.93...that's what i got before but i thought it would be DE not how fast DE is moving. so the answer to the question is -36.93? then why would my teacher give us the pythagorian theorem as a hint?
 
oh wait. i take that answer back. i forgot to take the derivative of 12+480/gt^2. After substituting -9.81 for g and find 1 for t and finding the derivative i got 97.9 m/s. Is this finally right?
 
Positive?

You didn't answer my question: This does NOT work for f(0) = 20 OR for f(t<sub>splat</sub>) = 0. You tell me why this is so and why it is important.
 
it's important because the speed is not the same at all times. the speed of the shadow changes with time. It is not constant.
 
You're getting there. I actually like the first try better.

At t = 0, the triangles don't exist. The shadow is out in space, somewhere. DE is infinite.
At f(t) = 0, the triangles don't exist. DE = 0
Domain considerations are very important.

There is a bit of a paradox around f(t) = 0.
If the speed, S, is 0, at some time, tI, and
S > 0 at some time before that, t1, such that t1 < tI, by continuity,
there must be some time, t2, such that tI > t2 > t1, at which the speed is S/2.
Further, there must be some other time, t3, such that tI > t3 > t2, at which the speed is S/4.
When does it EVER get to zero?

There is a related paradox around t = 0, but it's a little harder to fathom.
 
It gets to 0 when it stops or lands on the ground. I think i understand that. But I don't understand why the answer is not
DE(t)=12+(480/(-9.81 (t)^2))
DE'(1)=97.9 m/s
Aren't I looking for the derivative of side DE?
 
I lost track. Seems like it should be negative, but I don't see where I missed a sign. Perhaps another will straighten me out.
 
nevermind the last reply. I see now. But wouldn't it be f(t) = positive
1/2gt^2?
 
Whoops!

DE(t) = 12 + 480/[g*t<sup>2</sup>]

Not good. I was having trouble remembering that g < 0. That should be:

DE(t) = -12 - 480/[g*t<sup>2</sup>]

DE'(t) = 960/[g*t<sup>2</sup>] -- Remember that g < 0 so this is negative, as expected.

DE'(1) = 97.859 -- Ummm, what are the units? I have forgotten. m/s?

Ack! I did it again! Next time, I'm using g > 0 and adding the minus sign to the calculation. That's:

DE'(1) = -97.859 m/s
 
Thanks so much for your help. i learned a lot from this problem. thankyou again.
 
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