Height of cliff above water

  • Thread starter aksarb
  • Start date
  • #1
12
0

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?
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,819
14
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
12
0
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
mgb_phys
Science Advisor
Homework Helper
7,819
14
Yes (and some more unnecessary text to make the anwer longer than 4 characters otherwise the forum software rejects it )
 
  • #5
12
0
Thanks alot this si a great help. Do you think you can help me with the other question I have posted?
 

Related Threads on Height of cliff above water

  • Last Post
Replies
1
Views
1K
Replies
1
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
10K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
16K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
Top