Question

A ball is thrown straight down from the top of a 300-ft building with an initial velocity of -12ft per second. The position function is s(t)=-16t^2+v0t+s0. What is the velocity of the ball after 4 seconds?

Answer 1

\[s(t)=-16t^2+v_0t+s_0\]
v0= initial velocity = -12
s0 initial height = 300

Answer 2

answer would be -140?

Answer 3

????????

Answer 4

\[s(t)=-16t^2-12t+300\]

Answer 5

plug in 4 \for t and i am getting negative 4 as the answer

Answer 6

s'(t)=-32t-12
= -32(4)-12
=-140

Answer 7

sorry velocity ... yes -140 .