A 754 N diver drops from a board 9.00 m above the water's surface. The acceleration of gravity is 9.81 m/s^2.A.) Find the diver's speed 5.80 m...

It is assumed that in part A) and part B) of the question the the diver leaved the diving board with speed of 0 (zero).

A)

Total height of the board is 9.0 m above water surface. Therefore, when the diver is 5.8 meters above the water surface the total distance (s) covered by the diver due to acceleration due to gravity (g) is given by

s = 9.0 - 5.8 = 3.2 m

The velocity v when the diver has covered distance s is given by:

s = (g*s)^1/2 = (9.81*3.2)^1/2 = 5.6028 m/s

B)

Just before striking the water surface the total distance (s) covered by the diver due to acceleration due to gravity (g) is 9.0 m.

The velocity v when the diver has covered distance s is given by:

s = (g*s)^1/2 = (9.81*9.0)^1/2 = 9.3963 m/s

C) The upward distance travelled by diver before reaching velocity of 0 and starting to fall again is given by:

s1 = (u^2)/2g

Where u = initial upward speed = 2.9 m/s.

Therefore: s1 = (u^2)/2g = (2.9^2)/2.9.81 = 0.42864 m

Total distance (s) covered by diver under the influence of gravity from highest point to the surface of water is given by:

s = s1 + (Height of the board above water surface) = 0.42864 + 9.0 = 9.42864 m.

The velocity v when the diver has covered distance s is given by:

s = (g*s)^1/2 = (9.81*9.42864)^1/2 = 9.6174 m/s

Answer:

A)  5.6028 m/s

B)  9.3963 m/s

C)  9.6174 m/s