We calculate this by considering the change in the potential energy and kinetic energy from point B to point D.

By the law of the conservation of energy PE + KE remains unchanged.

At B, PE = 0 as the skater has nowhere to fall and KE = 1/2mv1^2, where m is the mass of the skater and v1 is the speed at B = 9.5m/s.

At D, PE = mgh, where g is the acceleration due to gravity and h is the height of D above the ground, and KE = 1/2mv2^2, where v2 is the speed at D which we wish to calculate.

Now, the angle the line OD makes with the horizontal is 30 degrees, so

h = rsin30 + r = 2sin30 + 2 = 3.

Therefore,

1/2mv2^2 + 3mg = 1/2mv1^2

implies v2^2 + 6g = v1^2

implies v2 = sqrt(v1^2 - 6g) = sqrt(9.5^2 - 6(9.8)) = 5.6 m/s

**The speed at D is 5.6m/s**