# Thread: calculus, finding velecity of shadow

1. ## calculus, finding velecity of shadow

here is the question:

a light source is dropped from a $\displaystyle \frac{1}{2}gT^2$ height tower top (t=0).
from a point which is $\displaystyle a$ distant from the tower base, a stone is thrown with velocity of $\displaystyle gT$ 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 $\displaystyle \frac{1}{2}gT^2$ height tower top (t=0).
from a point which is $\displaystyle a$ distant from the tower base, a stone is thrown with velocity of $\displaystyle gT$ 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, $\displaystyle y_L$, above the ground of the light source at time $\displaystyle t$.

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

• 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, $\displaystyle x$, of the shadow from the foot of the tower, in terms of $\displaystyle y_L, y_S$, and hence in terms of $\displaystyle t$.

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

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

I'll give you one or two steps:

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

(2) $\displaystyle y_S = ...$ ?

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

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

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

3. Hello, stud_02!

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

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

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

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

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

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

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

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

We have: $\displaystyle EC = a$

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

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

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

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

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

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

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

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

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.

By the way grandad your algebra is perfect.