Pre-calculus - Deriving position from acceleration

Question

Suppose an object is dropped from the tenth floor of a building whose roof is 50 feet above the point of release. Derive the formula for the position of the object t seconds after its release is distance is measured from the roof. The positive direction of distance is downward and time is measured from the instant the object is dropped. What is the distance fallen by the object in t seconds?

The problem gives you acceleration as being $a=32$. I integrate to velocity, getting $v=32t + C$. At the time of release, $v=0$, so the equation is $v=32t$. Integrate again for position and I get $$s=16t^2 + C$$ Here is where I get stuck. Do I add 50 to the position function because the height is measured from the roof? What is the next step?

Answer 1

Yes, add $50$ft to the function, because when $t=0$, the object is $50$ft away from the roof.

Answer 2

Also the second time you integrate you get

$$s = 16t^2 + Ct + C_2$$ - you get a new constant for each new integration. So it is this $C_2$ that you can help determine from the initial condition ( height at $t = 0$ ).

A monomial $t^n \rightarrow \frac{t^{n+1}}{n+1}$ under differentiation. So if $n = 1$ you see we get $$t^1 \rightarrow \frac{t^{1+1}}{1+1} = \frac{t^2}{2}$$ and for $n=0$ we get: $$t^0 \rightarrow \frac{t^{0+1}}{0+1} = t$$