# Finding the time when two stones cross paths

**1. The problem statement, all variables and given/known data**

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?

**2. Relevant equations**

d = vt + 0.5at^{2}

**3. The attempt at a solution**

distance travelled by both rock A and rock B combined is 6.69

D_{A} + D_{B} = 6.69

let the upwards direction be positive

D_{A}: (top of cliff)

d = v_{i}t + 0.5at^{2}

= (-9.97 m/s[up])t + 0.5(-9.8 m/s^{2}[up])t^{2}

= -9.97t – 4.9t^{2}

D_{B}: (base of cliff)

d = v_{i}t + 0.5at^{2}

d = (9.97 m/s [up])t + 0.5(-9.8m/s^{2}[up])t^{2}

d = 9.97t – 4.9t^{2}

D_{A} + D_{B} = 6.69

-9.97t – 4.9t^{2} + 9.97t – 4.9t^{2} = 6.69

-9.8t^{2} = 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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