Question

A stone is thrown vertically downward from a cliff 186.0-m tall. During the last half second of its flight, the stone travels a distance of 46.9 m. Find the initial speed of the stone. Note: I won't be able to respond for a few hours

Answer 1

Is this just basic distance time speed formula?
If so...piece of cake. :D

Answer 2

Nm

Answer 3

They don't give a time

Answer 4

crapola

Answer 5

back to the drawing board then

Answer 6

figured it out
I think

Answer 7

divide 186 by 46.9

Answer 8

multiply the result by .5

Answer 9

1.98

Answer 10

1.98 in minutes and seconds is...

Answer 11

1 minute and 58 seconds

Answer 12

I THINK

I'm no physicist...

Answer 13

Speaking of which, I must admit @Abmon98, I MIGHT be stalking you for help in Chemistry

Answer 14

wait, wait, wait
I read it as half a minute, not half a second
crapola

Answer 15

haahah sure i will help you give me 5 more minutes

Answer 16

well...that would just mean...
1 second and 3 half seconds

and thanks @Abmon98

Answer 17

s=46.9 u=xm/s a=10m/s^2 t=0.5 second
ut+1/2at^2=s
that initial velocity is the final velocity when the stone was travelling ((186-46.9)=139.1m
at s=139.1 m a=10m/s^2 v=91.300 m/s
use Vt-1/2at^2=s
solve for t and then use u+v/2)*t=s