- #1

aksarb

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## Homework Statement

You throw a rock off a cliff, giving it a velocity of 8.3 m/s straight down. At the instant you released the rock, your hiking buddy started a stopwatch. The rock hit the water below exactly 6.9 seconds after you threw the rock. How high is the cliff above the water?

## Homework Equations

Δd =

__1__(v1+v2)Δt

2

2aΔd = v2^2 - v1^2

Δd = v1 Δt + 1/2 a (Δt)2

Δd = v2 Δt - 1/2 a (Δt)2

## The Attempt at a Solution

I am assuming that the v1 is 0 because it starts from rest and v2 is -8.3m/s. (Sign convention was down negative.) and time was 6.9. I subbed everything into the equation. When I make everything equal to Δd, does that give me the height of the cliff above the river?