Conservation of Energy confusion

[kisbester]

Homework Statement

A dog takes a running horizontal leap off a 10 m cliff and jumps with a speed of 3 m/s onto a ledge 4 m below the height of the cliff. With what speed does he land on the ledge?

Homework Equations

KE_i + PE_i = KE_f + PE_f
1/2mv_i² + mgh_i = 1/2mv_f² + mgh_f

The Attempt at a Solution

Attempt #1:

(masses cancel, g ≈ 10m/s²)

1/2(3m/s)² + (10m/s²)(10m) = 1/2v_f² + (10m/s²)(4m)

1/2(9)+100 = 1/2v_f² + 40

4.5 + 60 = 1/2v_f²

2(64.5) = 129 = v_f²

\sqrt{}129 = v_f

v_f = 11.36

This is not the correct answer, however, upon looking at the solution the only difference is that they make the original height equal to zero and the final height equal to -4. This of course excludes PE from the first half of the equation. I realize that it would have been simpler for me to do the problem in that way. I don't, however, understand why it doesn't work to do it the way that I did it. Shouldn't it only be the difference in PE that matters? Should the final height have been something different than 4 for my version to work? Any clarification would be greatly appreciated.

 

[Doc Al]

This is not the correct answer, however, upon looking at the solution the only difference is that they make the original height equal to zero and the final height equal to -4. This of course excludes PE from the first half of the equation. I realize that it would have been simpler for me to do the problem in that way. I don't, however, understand why it doesn't work to do it the way that I did it. Shouldn't it only be the difference in PE that matters? Should the final height have been something different than 4 for my version to work? Any clarification would be greatly appreciated.

Your error is thinking that the height of the ledge is 4 m. Note that the ledge is stated to be 4 m below the cliff, not 4 m from the bottom.

If you compare your calculation to the solution given, you'll see that the difference in PE is not the same.

Get the correct height of the ledge and your method is fine.

 

[kisbester]

Ugh, thank you so much.