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!*

.