A ball is thrown straight down from the top of a 280 ft building with an initial velocity of -22 ft per second. The position function is s(t)=-16t^(2)+v_(o)t+s_(o)

Question

anonymous · Answer

I don't know how to write in subscript so I did v_(o) = inital velocity and s_(o)= initial position.

anonymous · Answer

$$s(t) = V_0 + {1 \over 2} \times g \times t^2 ?$$
what's with the 16?

anonymous · Answer

Oops, i forgot, you're not metric?

anonymous · Answer

Sorry, you have to find the velocity of the ball after 2 seconds.  & haha, yeah.

anonymous · Answer

Initial speed -11 f/s. after 1 second (-11 -16 ) = -27. After 2 seconds (-27-32)=-59 f/s

anonymous · Answer

oops, initial -22, that makes -70 f/s then

anonymous · Answer

You don't really need the position formula there, maybe just to check it didnt actually hit the ground before 2 seconds have passed.

anonymous · Answer

Thank youuu(: