Step 1 :The position of an object dropped from a 172-foot-high building is given by \(s(t)=-16 t^{2}+172\), where \(t\) is the time in seconds since it was dropped.
Step 2 :To find the velocity of the object 1 second after being dropped, we need to find the derivative of \(s(t)\), which represents the velocity of the object at any given time \(t\).
Step 3 :The derivative of \(s(t)=-16 t^{2}+172\) is \(s'(t)=-32t\).
Step 4 :Evaluating this at \(t=1\) will give us the velocity 1 second after being dropped.
Step 5 :Substituting \(t=1\) into \(s'(t)=-32t\), we get \(s'(1)=-32(1)=-32\).
Step 6 :Final Answer: The object's velocity 1 second after being dropped is \(\boxed{-32 \frac{\mathrm{ft}}{\mathrm{sec}}}\).