Question

a stone is thrown vertically upward with a speed of 12.5m/s fro the edge of a cliff 75.0m high. How much later does it hit the bottom of the cliff, what is the speed just before hitting? what total distance did it travel?

Answer 1

i know that: Vi=12.5m/s, Xi=75m, X=-75m, a=g=9.8m/s^2 but i don't know which equations to use to find time,final velocity and total displacement :/

Answer 2

Xf=0m
a=-g=-9.8m/s^2

Answer 3

okay, so a=-g for the 1st half of the motion when it is thrown upward then afterwards when its going down its a=g because it going down since the ball was thrown at the edge of a cliff..

Answer 4

Now use \[x _{f}-x_{i}=v_{i}t+at^2/2\]

Answer 5

No, it is always -g

Answer 6

how come?

Answer 7

You see, you are specifying a coordinate system in which the bottom of the cliff is 0, upward direction is +X & the downward direction is -X
g=down=negative

Answer 8

the direction of acceleration never changes. It is only the direction of the velocity that changes.
v=>up=+ve
=>down=-ve

Answer 9

oh okay (: thx

Answer 10

ah! i got it! t=5.39s^^

Answer 11

do i use v=vi+at to find the velocity before it hit the bottom?

Answer 12

yes

Answer 13

but i got t=2.2s

Answer 14

oh wait..i know what i did wrong now my xi and xf where mixed =.=

Answer 15

no , t=3.48s. Sorry about that.

Answer 16

its k so v=12.5+(-9.8 x3.48)

Answer 17

yep.

Answer 18

and total displacement will just be rearranging v^2=vi^2+2as to solve for s?

Answer 19

you got it!

Answer 20

^^ thanks to you!

Answer 21

Ahh.. but wait. I was blind

Answer 22

lol okay..

Answer 23

total distance traveled is what is asked.

Answer 24

oh okay so how do i find it then? :s

Answer 25

for that, you have to calculate how high (from the edge of the cliff) the stone traveled.

Answer 26

okay then add it to the height of the cliff

Answer 27

if that is 's',
total distance traveled=2s+75

Answer 28

oh okay :) thank you!

Answer 29

yw:)

Answer 30

2s=v^2-vi^2/a right?

Answer 31

No, it is