Question

A ball is thrown straight down from the top of a 449-foot building with an initial velocity of -15 feet per second. Use the position function below for free-falling objects.
s(t) = -16t2 + v0t + s0
What is its velocity after 2 seconds?
v(2) =
-47
ft/s

What is its velocity after falling 316 feet?
v =
ft/s

Answer 1

old question but will respond so this can be closed

they give you: s(t) = -16t^2 + v0_t + s0
where s(t) is the position at (t) seconds, v_0 is the initial velocity (given as -15 ft/s), s0 is the initial position (449 ft)

for the first Q: What is its velocity after 2 seconds? ---> plug in t = 2
for the second Q: this is a little trickier. it fell 316 feet, so the *change* in position s(t) - s(t0) equals 316

in terms of the equation, it looks like this:
s(t) = -16t^2 + v0_t + s0
subtracting s0 from both sides

s(t) - s0 = -16t^2 + v0_t
so on the left side, let s(t) - s0 = 316, v_0 stays the same, and solve for t