Posted by **Tally** on Thursday, August 11, 2011 at 10:02am.

A person dives off the edge of a cliff 40m above the surface of the sea below. Assuming that air resistance is negligible, (a) How long does it dive last? (b) and with what speed does the person enter the water?

- physical science 101 -
**Henry**, Thursday, August 11, 2011 at 7:04pm
a. d = Vo*t + 0.5at^2 = 40m,

0 + 0.5 * 9.8 * t^2 = 40,

4.9t^2 = 40,

t^2 = 8.16,

t = 2.86s.

b. Vf^2 = Vo^2 + 2gd,

Vf^2 = 0 + 2 * 9.8 *40 = 784,

Vf = 28m/s.

- physical science 101 -
**Jessica**, Sunday, January 29, 2012 at 7:55pm
A) h=0.5gt^2

40=0.5(9.8)(t^2)

40=4.9t^2

t^2=8.16

t=2.86s

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