Question

A ball is thrown straight down from the top of a 220 foot building with an initial velocity of -22 ft/sec. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?

Answer 1

devise en equation; and derive it

Answer 2

h(t) = -Gt^2 + Vi t + Hi

Answer 3

and since we are in feet, G = 16, and they give you Vi as -22 and Hi as 220

Answer 4

I figured out how to get the velocity by finding the derivative of -16t^2-22t+220. But how do you get the velocity after falling 108 feet?

Answer 5

equate h(t) to 108 and solve for "t"

Answer 6

h(t) = -16t^2-22t+220 = 108

Answer 7

remember the quadratic formula if anything, it tends to be the most plug and play for it

Answer 8

i cant quite be sure if deriving the new equation will suffice; it might .... but i cant tell if it does or not

Answer 9

id find "t" from it and go back to the derivative to begin with

Answer 10

with any luck, that makes sense ;)

Answer 11

-16t^2-22t+220 = 108 /-2

8t^2 +11t -110 = -54

8t^2 +11t -110 = -54
+54 +54
----------------------
8t^2 +11t - 56 = 0

t = \(\cfrac{-11 \pm \sqrt{121-4(8)(-56)}}{2(8)}\)


and disregard the "bad" result