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## 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?

## Homework Equations

d = vt + 0.5at

^{2}

## 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?