
#1
Oct707, 12:34 PM

P: 12

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? 



#2
Oct707, 12:39 PM

Sci Advisor
HW Helper
P: 8,961

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. 



#3
Oct707, 12:47 PM

P: 12

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? 



#4
Oct707, 12:48 PM

Sci Advisor
HW Helper
P: 8,961

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




#5
Oct707, 12:51 PM

P: 12

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



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