# Finding the time when two stones cross paths

## Homework Statement

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.69 m. The stones are thrown with the same speed of 9.97 m/s. After what time do the stones cross paths?

d = vt + 0.5at2

## The Attempt at a Solution

distance travelled by both rock A and rock B combined is 6.69
DA + DB = 6.69

let the upwards direction be positive

DA: (top of cliff)

d = vit + 0.5at2
= (-9.97 m/s[up])t + 0.5(-9.8 m/s2[up])t2
= -9.97t - 4.9t2

DB: (base of cliff)
d = vit + 0.5at2
d = (9.97 m/s [up])t + 0.5(-9.8m/s2[up])t2
d = 9.97t - 4.9t2

DA + DB = 6.69
-9.97t - 4.9t2 + 9.97t - 4.9t2 = 6.69
-9.8t2 = 6.69 getting negative time...?
t = 0.82 s

the correct answer is 0.335 seconds. I think I am messing up my directions and signs for my vectors... any help?

Good start.
Three things…

1. Keep your numbers as variables. It's alot easier to keep writing "g" than to keep writing "-9.8 m/s2[up]".
2. You said.

"distance travelled by both rock A and rock B combined is 6.69
DA + DB = 6.69"

THis is close, but not quite correct. We don't know that this is the case. It would work if A was going down the whole time and B was going up the whole time. We know A will be going down the whole time, but B will reach a maximum height and begin to fall.

Your equation should mathematically say " The height of the cliff is 6.69 m".

1 person
Missed my third point…

"-9.8t2 = 6.69 getting negative time…?"

Nope, you wouldn't get negative time. It's even weirder; You would have imaginary time!

Dick
Homework Helper
It's probably easier to think about if you pick a common origin for ##D_A## and ##D_B##. Pick the common origin ##D=0## at the base of the cliff. So ##D_A=-9.97t - 4.9t^2 + 6.69## and ##D_B=9.97t-4.9t^2##. Now just solve ##D_A=D_B##.

1 person
It's probably easier to think about if you pick a common origin for ##D_A## and ##D_B##. Pick the common origin ##D=0## at the base of the cliff. So ##D_A=-9.97t - 4.9t^2 + 6.69## and ##D_B=9.97t-4.9t^2##. Now just solve ##D_A=D_B##.

##D_A + 6.69 = D_B##

It took me a while, but I finally found meaning behind this equation...
displacement of rock A: ##D_A = -6.69 m [up]##, and B: ##D_B = 0 m.##

-6.69 + 6.69 = 0, brilliant!