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 + 0.5at2 3. 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?