- #1

- 2

- 0

**Object dropped from rest at h**

_{0}meters. At what height, h_{1}, will K=2U? g=9.8m/s^{2}**K=2U**

E

E

_{T}=K-U**(.5)mv**

v

(v

^{2}=2(mgh)v

^{2}=gh(v

^{2})/g=hThanks. :)

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Noms
- Start date

- #1

- 2

- 0

E

v

(v

Thanks. :)

- #2

Andrew Mason

Science Advisor

Homework Helper

- 7,680

- 401

Your question does not make sense. Please give us the entire question exactly as you were given it.Object dropped from rest at h_{0}meters. At what height, h_{1}, will K=2U? g=9.8m/s^{2}

K=2U

E_{T}=K-U

(.5)mv^{2}=2(mgh)

v^{2}=gh

(v^{2})/g=h

Thanks. :)

AM

- #3

- 2

- 0

After falling, the total energy will be E' = K' + U' = 2U' + U' = 3U'

As we know, E = E'

Then, mgh = 3mgh' --> h' = 1/3 h

- #4

- 2

- 0

M- That's just what I needed. :)

Share:

- Replies
- 1

- Views
- 6K