# 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?

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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".

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 $D_A=D_B$.

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!