Question

pls help!

98. A ball is thrown straight down from the top of a 220-foot
building with an initial velocity of -22 feet per second. What
is its velocity after 3 seconds? What is its velocity after falling108 feet?

position function=-16t^2+v0t+s0
i think the answer to the first part is -118 ft/second and im lost on the second part

Answer 1

take the derivative to determine velocity

Answer 2

yea so you would get v(t)=-32t-22 and then you plug in 3 and get -118 correct? what about the second part

Answer 3

i edited the question sorry it was messed up

Answer 4

s(t) = -16t^2-22t+220

s'(t) = v(t) = -32t-22, yep. velocity gives speed and direction which tells us that its -118 downward

Answer 5

solve for t when s(t) = 220-108

Answer 6

then we can use velocity againt

Answer 7

so i used the quadratic equation and i got 2, and -27/8

Answer 8

so now what

Answer 9

well, since time tends to start at 0 and just grows from there, id let t=2

Answer 10

v(2) = -32(2) - 22