A ball is thrown straight down from the top of a 210-ft building with an initial velocity of -24 ft. per second. Find the velocity of an object after it has fallen 130 feet.

Question

A ball is thrown straight down from the top of a 210-ft building with an initial velocity of -24 ft. per second.  Find the velocity of an object after it has fallen 130 feet.

moonlitfate · Answer

Just to make things simpler, I know the position function for the ball is -16t^2-24t+210

And the velocity function is the derivative of that: -32t-24.  Just  not sure where to go from there.

anonymous · Answer

66

anonymous · Answer

98

anonymous · Answer

jk

anonymous · Answer

idn sorry

anonymous · Answer

they should have gave a graph?

moonlitfate · Answer

Can I actually get some useful help from someone out there. ._.

anonymous · Answer

set it equal to 180 and solve, then you know what the time is, plug it in to the derivative

anonymous · Answer

that was wrong, sorry, set it equal to 80 since if it falls 130 feet it is 80 feet up
then solve for $t$ and plug that in to the derivative

moonlitfate · Answer

@satellite73 -- I'm definitely doing something wrong here. ._.; 

I've set s(t)=-16t^2+24t = 80 and solved for t then plugged in back into the derivative and got an answer of -42.97,  but it isn't right.  The answer should be -94.32.