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?

Answer 1

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.

Answer 2

A) h=0.5gt^2
40=0.5(9.8)(t^2)
40=4.9t^2
t^2=8.16
t=2.86s