A woman 6 ft tall walks away from a 17 ft tall pole with a street light on top of it.
Graphically:
From the diagram, we can say that:
[tex]\frac{x}{6}=\frac{y}{11}\ldots(1)[/tex]The laws of motion of the woman and the shadow with respect to the base of the pole are:
[tex]\begin{gathered} y(t)=y_0+v_y\cdot t \\ x(t)=x_0+v_x\cdot t \end{gathered}[/tex]Where x0 and y0 are the initial positions, vx and vy are the shadow speed and the speed of the woman, respectively. We know that vy = 4 ft/s, so if we say that the starting point for the woman was 11 feet away from the pole, then using equation (1):
[tex]\begin{gathered} y_0=11 \\ \frac{x_0}{6}=\frac{y_0}{11}\Rightarrow x_0=6 \end{gathered}[/tex]Then, the laws of motion are:
[tex]\begin{gathered} y(t)=11+4\cdot t \\ x(t)=6+v_x\cdot t \end{gathered}[/tex]We now calculate the instant of time when y = 45 ft. Using the law of motion of the woman:
[tex]\begin{gathered} 45=11+4\cdot t \\ 34=4\cdot t \\ t=8.5\text{ sec} \end{gathered}[/tex]Again, using the equation (1):
[tex]\begin{gathered} \frac{x(8.5)}{6}=\frac{y(8.5)}{11} \\ x(8.5)=\frac{6\cdot45}{11} \\ 6+v_x\cdot8.5=\frac{270}{11} \\ 8.5v_x=\frac{204}{11} \\ v_x=\frac{24}{11}\text{ ft/sec} \end{gathered}[/tex]
