A.) If the diver leaves the board with an initial upward speed of 2.90 m/s, find the diver's speed when striking the water. Answer in units of...

a)

The motion of the diver upward with a initial velocity of 2.9m/s.

Taking the motion directed upward as postive, the equation of mtion of the diver with initial speed u, and displecemnt s intime t is given by

s=ut-(1/2)gt^2., wher g is the gravitational acceleration and is equal to 9.81m/s^2 and u =2.9.

When the diver reches the water below 9m ,  s= -9m So the equation becomes:

-9 = 2.9*t-0.5*9.81t^2 0r

4.905t^2-2.9t-1.3864539 = 0.  We solve for t, time taken by the diver to reach the water.

t =[ 2.9 + sqrt(2.9^2+4*4.905*9)]/(2*4.905)

=  1.68207s

Therefore, the speed when he strike water = u-gt =2.9-9.81(1.68207)= -13.6011m/s

b)

The divers speed at 5.5 m above the water surface . So s = -3.5 m below the board.

-3.5= 2.9t- 4.905^2 or

4.905t^2-2.9t-3.5=0

t = 1.19057 s

v=2.9-9.81(1.19057) = -8.7795 m/s