1. ## calculus, finding velecity of shadow

here is the question:

a light source is dropped from a $\frac{1}{2}gT^2$ height tower top (t=0).
from a point which is $a$ distant from the tower base, a stone is thrown with velocity of $gT$ vertically at the same time.
find the velocity of the stone's shadow when t=T/4

i can find velocity of the light source of the light source and stone at a t=t.
but can't find the velocity of the shadow using them.

2. ## Displacement and velocity

Hello stud_02
Originally Posted by stud_02
here is the question:

a light source is dropped from a $\frac{1}{2}gT^2$ height tower top (t=0).
from a point which is $a$ distant from the tower base, a stone is thrown with velocity of $gT$ vertically at the same time.
find the velocity of the stone's shadow when t=T/4

i can find velocity of the light source of the light source and stone at a t=t.
but can't find the velocity of the shadow using them.
I assume that the shadow is cast on the ground, which is horizontal.

Then the way to do this is as follows:

• Use the equations of constant acceleration to find an expression for the height, $y_L$, above the ground of the light source at time $t$.

• Find an expression for the height, $y_S$, above the ground of the stone at time $t$.

• Draw a straight line from the light source to the stone, and produce it until it meets the ground. The point where it meets the ground is the position of the shadow.

• Use similar triangles to find the distance, $x$, of the shadow from the foot of the tower, in terms of $y_L, y_S$, and hence in terms of $t$.

• Differentiate $x$ with respect to $t$ to get an expression for $\dot{x}$, the velocity of the shadow in terms of $t$.

• Put $t = \tfrac14T$ to find the answer.

I'll give you one or two steps:

(1) $y_L = \tfrac12gT^2 - \tfrac12gt^2$

(2) $y_S = ...$ ?

(3) $x = \frac{ay_L}{y_L-y_S}=...$ ?

(4) $\dot{x} = ...$ ?

If I haven't made any mistakes with the algebra, I think the answer is $\frac{13a}{2T}$.

3. Hello, stud_02!

Wow! . . . What a long, intricate problem!

A light source is dropped from a $\tfrac{1}{2}gT^2$ foot tower top (t=0).

From a point which is $a$ distant from the tower base,
a stone is thrown with velocity of $gT$ vertically at the same time.

Find the velocity of the stone's shadow when $t=\tfrac{T}{4}$
Code:
                              A
*   -   -
|   :   :
|   :  16gt²
|   :   :
* B :   -
*  |   :   :
*     | ½gT²  :
D  *        |   :   :
o         y |   :  16gT²-16gt²
*  |           |   :   :
*     |gTt        |   :   :
*        |           |   :   :
o-----------*-----------*   -   -
S    x-a    E     a     C
: - - - - - x - - - - - :

The tower is: $AC = \tfrac{1}{2}gT^2$

In $t$ seconds, the light falls: $AB = 16gt^2$ feet.

Hence: . $y \:=\:BC \:=\:16gT^2 - 16gt^2\;\;{\color{blue}[1]}$

The stone is thrown upward from $E\!:\;DE = gTt$ feet.

We have: $EC = a$

The shadow of the stone is at $S\!:\;\text{ let }x = SC.$
. . Then: $SE \:=\: x-a$

We have similar right triangles: . $\Delta BCS \sim \Delta DES$

So we have: . $\frac{x}{y} \:=\:\frac{x-a}{gTt} \quad\Rightarrow\quad x \:=\:a\cdot\frac{y}{y-gTt}$

Substitute [1]: . $x \;=\;a\cdot\frac{\frac{1}{2}g(T\:\!^2-t^2)}{\frac{1}{2}g(T\:\!^2-t^2) - gTt} \;=\;\frac{\frac{1}{2}g(T\:\!^2-t^2)}{\frac{1}{2}g(T\:\!^2-t^2 - 2Tt)}$

. . and we have: . $x \;=\;a\cdot\frac{T\:\!^2-t^2}{T\:\!^2-t^2-2Tt}$

Differentiate: . $\frac{dx}{dt} \;=\;a\cdot\frac{(T\:\!^2 - t^2-2Tt)(-2t) - (T\:\!^2-t^2)(-2t-2T)}{(T\:\!^2-t^2-2Tt)^2}$

. . which simplifies to: . $\frac{dx}{dt} \;=\;\frac{2aT(T\:\!^2 + t^2)}{(T\:\!^2 + t^2 - 2Tt)^2}$

Now plug in $t = \frac{T}{4}$

But check my work . . .please!
.

4. ## Animation

Excellent problem.

To all concerned-- I'll be adding an animation on my website
Calculus Animations

later today demonstrating this problem.

Just go to the related rates page and look for falling light source.