Answer:
v = 6.95 m/s
Explanation:
Given that,
A diver is on a board 1.80 m above the water, s = 1.8 m
The initial speed of the diver, u = 3.62 m/s
Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :
[tex]v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s[/tex]
So, she will hit the water with a speed of 6.95 m/s.