Answer
$-1500ft/sec$
Work Step by Step
Step 1. Using the figure given in the Exercise, we are given $h=50ft, x_1=30ft, t=0.5sec$ and $s=16t^2$
Step 2. At the time after 0.5sec, the height of the ball is $y=50-16(0.5)^2=46ft$
Step 3. Position of the shadow $x(t)$ satisfies: $\frac{x-30}{x}=\frac{y}{50}$ (similar triangles), which gives $xy=50x-1500$ or $(50-y)x=1500$ and $x=\frac{1500}{50-y}$
Step 4. The shadow will move at a velocity $\frac{dx}{dt}=\frac{1500}{(50-y)^2}\frac{dy}{dt}$
Step 5. From the equation $s=16t^2$, we have $\frac{dy}{dt}=-\frac{ds}{dt}=-32t=-32(0.5)=-16ft/s$
Step 6. Finally, $\frac{dx}{dt}=\frac{1500}{(50-46)^2}(-16)=-1500ft/sec$
