Height of cliff above water

aksarb

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

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?

2. Relevant 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

3. 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?

mgb_phys

All you need is the s = ut + 1/2at^2
But the initial velocity u=8.3 and a=g=9.8m/s^2
Don't try and remember the sign convention, think about would S get bigger or smaller with time, would a large U reduce/increase S.

aksarb

So, it would be:

D = (8.3m/s)(6.9s) + 1/2 (9.80/ms^2)(6.9s)^2

Work it out and the D would give me the height?

mgb_phys

Yes (and some more unnecessary text to make the anwer longer than 4 characters otherwise the forum software rejects it )

aksarb

Thanks alot this si a great help. Do you think you can help me with the other question I have posted?