A ball is thrown straight down from the top of a 445-foot building with an initial velocity of -24 feet per second. Use the position function below for free-falling objects.

Question

s(t) = -16t2 + v0t + s0

What is its velocity after 2 seconds?
V(2)= _____ ft/s

What is its velocity after falling 216 feet?
V= _____ ft/s

Answer 1

So your equation would be
s(t) = -16t^2 - 24y + 445
v(t) = -32t - 24

so v(2) = ..... just sub in t = 2

after falling 216 ft, the height is still 229 ft

so solve
-16t^2 - 24t + 455 = 229 for t, then sub it into v(t)
(reject any negative t , is they occur)

Answer 2

A ball is thrown straight down from the top of a 500-foot building with an initial velocity of -19 feet per second. Use the position function below for free-falling objects.

s(t) = -16t2 + v0t + s0

What is its velocity after 3 seconds?
V(2)= _____ ft/s

What is its velocity after falling 332 feet?
V= _____ ft/s