If an object is dropped from a 172-foot-high building, its position (in feet above the ground) is given by $s(t)=-16 t^{2}+172$, where $t$ is the time in seconds since it was dropped.
(a) What is its velocity 1 second after being dropped?
(b) When will it hit the ground? (Hint: Solve $s(t)=0$.)
(c) What is its velocity upon impact (at the instant when it hits the ground)?

The object's velocity 1 second after being dropped is $\square \frac{\mathrm{ft}}{\mathrm{sec}}$.
(Simplify your answer.)

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 \(\boxed{-32 \frac{\mathrm{ft}}{\mathrm{sec}}}\).