Homework Help: physical science 101

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

Answer this Question

Related Questions

Physical Science - A person dives off the edge of a cliff 33m above the surface ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...
Physics - A person standing at the edge of a seaside cliff kicks a stone ...
Physics - A person standing at the edge of a seaside cliff kicks a stone ...
science - a 110 kg person dives off a cliff and hits the water at 17 m/s. how ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...
physics - A person standing at the edge of a seaside cliff kicks a stone over ...

For Further Reading

First Name:
School Subject:
Answer: