Question 1153645
first draw it:

{{{drawing ( 600, 600, -10, 15, -10, 10,

circle(0,8,.14),circle(5,8,.12),
green(line(5,8,5,0)),circle(5,5.2,.12),circle(14,0,.12),green(line(0,8,14,0)),
locate(0.5,4,40ft),locate(2.5,0.5,25ft),locate(7.5,0.5,x-25ft),locate(5,5.6,(1/2)s-later),locate(5.3,3,40ft-s),line(0,-1,14,-1),locate(5,-0.5,x),
graph( 600, 600, -10, 15, -10, 10, 0)) }}}


given: 

a building wall that is {{{40ft}}} high
a tree house that is {{{25ft}}} to the right of the building
the shadow of the apple moving along the ground {{{1/2}}} sec later

{{{s=16𝑡^2}}} in {{{t}}} seconds 

two right triangles are similar, means corresponding sides are proportional

so,

{{{x/(x-24)=40/(40-s)}}}

{{{x(40-s)=40(x-24)}}}

{{{cross(40x)-xs=cross(40x)-960}}}

{{{-xs=-960}}}....multiply by {{{-1}}}

{{{xs=960}}}

{{{x=960/s}}}

Differentiating with respect to {{{t}}} we have

{{{dx/dt=-(960/s^2)(ds/dt)}}}


We are given {{{s=16t^2}}}⇒ {{{ds/dt = 32t}}}

 At {{{t = 1/2}}}, {{{s = 16(1/2)^2=16/4 = 4}}} and {{{ds/dt = 32/2 = 16}}}. 


Plugging these in we have

{{{dx/dt=-(960/4^2)(16)(ft/s)=-960(ft/s)}}}